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The gradient factors asIf a coordinate is zero, the solutions are $(0,0)$, $(\pm 1,0)$ and $(0,\pm 1)$. With both coordinates nonzero, subtracting the two bracket equations gives $x^2=y^2$, hence $x^2=y^2=1/4$. These give four more stationary points.

The Hessian matrix has equal diagonal entries $6xy$ and off-diagonal entry $3(x^2+y^2)-1$. At the origin its eigenvalues are $\pm 1$, and at the four nonzero axis points they are $\pm 2$, so all five are saddles. At the two same-sign interior points the eigenvalues are $1,2$, while at the opposite-sign points they are $-1,-2$. ThereforeThe extrema are not global, since the quartic takes arbitrarily large positive and negative values along suitable rays.

For the contour sketch, the zero level is exactly the two axes together with the unit circle. Inside the circle, $f$ is negative in the first and third quadrants and positive in the other two; outside, those signs reverse. Small negative levels contain closed loops around the two minima as well as unbounded branches in the second and fourth quadrants. Small positive levels similarly contain two loops around the maxima and unbounded branches in the first and third quadrants. The closed loops shrink to the extrema at levels $\mp 1/8$, and disappear beyond those levels. This is the unit-scale quartic surface with axes and a circle as its zero set.

The gradient equations factor asFor $y=0$ they give $x=\pm 5$. Otherwise $y=4x/3$, which gives $x=\pm 3$ and the matching $y=\pm 4$. Thus there are exactly four stationary points.

The Hessian matrix isAt $(5,0)$ it is $60I$, and at $(-5,0)$ it is $-60I$. At $(3,4)$ and $(-3,-4)$ its determinant is $-3600<0$. The second-derivative test therefore yieldsThe extrema are local: along $y=0$ the cubic is unbounded both above and below.

The partial derivatives give $x^3=9y$ and $y^3=9x$. If either coordinate is zero, both vanish. Otherwise substitution gives $x^8=3^8$, so the real stationary points are $(0,0)$, $(3,3)$ and $(-3,-3)$.

The Hessian matrix isAt the origin its eigenvalues are $36$ and $-36$, so it is a saddle at level zero. At either of the other points its eigenvalues are $72$ and $144$, so both are strict minima at level $-162$. In fact,with equality only at those two points. HenceFor levels $-162<c<0$, two separate closed contours surround the two minima. At $c=0$ the contours form two lobes, in the first and third quadrants, meeting at the origin. Their limiting tangent directions there are the coordinate axes because the leading term is $-36xy$. On the diagonal the outer zero-level points are $(\pm 3\sqrt{2},\pm 3\sqrt{2})$. Positive levels form one closed outer contour enclosing both wells; there are no contours below $-162$. The graph is symmetric under exchanging $x,y$ and under reversing both signs, but not under a quarter-turn. These features are the $a=3$ case of the quartic double well with bilinear coupling.

The partial derivatives are $g_x=\sin x$ and $g_y=y$, so all stationary points are $(n\pi ,0)$ for integers $n$. Their Hessian matrix is $\mathrm{diag}(\cos x,1)$. At even multiples of $\pi$ it is a positive-definite matrix, whereas at odd multiples it has one positive and one negative eigenvalue. ConsequentlyThere are no local maxima. Since both $1-\cos x$ and $y^2/2$ are nonnegative, these are also global minima. In the requested open plotting window, the stationary points are $(0,0)$ and $(\pm \pi ,0)$; the local minima at $(\pm 2\pi ,0)$ lie on excluded edges.

For a level set $g=c$, solving for the vertical coordinate givesFor $0<c<2$ these level sets are closed ovals around each local minimum. At $c=0$ the level sets shrink to the local minima. The level set through the saddle points iswhere the two signed branches together describe the same set as $y=\pm 2|\cos (x/2)|$. They cross at $(\pm \pi ,0)$ and reach $(0,\pm 2)$. The crossing slopes at a saddle point are $\pm 1$, as also follows from the local quadratic form $(-\delta x^2+\delta y^2)/2$. For $c>2$, the upper and lower level sets remain separate and extend across the plotting window without meeting the horizontal axis.

Use $\mathcal{L}=-{\sum }_i{x}_i\log x_i-\lambda ({\sum }_i{x}_i-1)$. Differentiating with respect to each positive $x_i$ givesEvery $x_i$ is therefore equal; normalization forcesThe Hessian matrix of $f$ is diagonal with entries $-1/x_i$, so its quadratic form is strictly negative for every nonzero perturbation. In particular it is negative along the tangent space ${\sum }_i\delta x_i=0$, making this the unique global maximum on the positive probability simplex. This is the maximum entropy distribution on a finite set, using natural logarithms.

The first partial derivatives factor asIf either coordinate vanishes, the other equation forces the other coordinate to vanish as well. If both are nonzero, subtracting the two bracket equations gives $x^2=y^2$, and then $x^2=y^2=1$. Thus the complete list of stationary points is $(0,0)$ and the four points $(\pm 1,\pm 1)$.

The Hessian matrix isAt the origin it is $8I$, which is a positive-definite matrix. At the four other points its eigenvalues are $32$ and $-16$, independently of the signs. The second-derivative test therefore givesThere are no local maxima. The minimum at the origin is not global, since $g(t,t)=8t^2-4t^4\to -\infty$ as $|t|\to \infty$.

The gradient vanishes at the origin. Its Hessian matrix there iswith eigenvalues $1$ and $-1$. The quadratic term is therefore indefinite, so the origin is a saddle point of a scalar function. Equivalently, the leading term $xy$ is positive along $y=x$ and negative along $y=-x$.

Put $G=(x^2+y^2)/2+a{x}^4/4$, so $f=e^G>0$. Since $f_x=f{G}_x$ and $f_y=f{G}_y$, its stationary points satisfyThe origin occurs for every real $a$. There are two additional points only for $a<0$, at $x=\pm 1/\sqrt{-a}$.

At a stationary point, the Hessian matrix simplifies because $\nabla G=0$:At the origin this is $I$, so $(0,0)$ is a strict local minimum for every $a$, with $f=1$. If $a\ge 0$ it is also the unique global minimum, since $G\ge 0$ with equality only there. For $a<0$, the two extra points have $1+3a{x}^2=-2$, so their Hessian has opposite-sign eigenvalues. ThusThere are no local maxima. When $a<0$, the origin is not a global minimum: $f(x,0)\to 0$ as $|x|\to \infty$, but $f$ never attains zero.

Evaluate the Hessian matrix at the five stationary points. At the origin it iswhose eigenvalues are $-6$ and $2$, so the origin is a saddle point. At the two points with $x=y=\pm \sqrt{3}/2$,whose eigenvalues are $12e^{-3/2}$ and $8e^{-3/2}$. Both are positive, so these are strict local minima. At the two points with $x=-y=\pm 1/2$,whose eigenvalues are $-8e^{-1/2}$ and $-4e^{-1/2}$. Both are negative, so these are strict local maxima. In summary,The quadratic changes along the orthogonal directions $(1,1)$ and $(1,-1)$ also provide a direct justification of these classifications.

Eliminate $z$ using the constraint: $z=x-x^2-y^2$. The restricted objective is the quadratic polynomial $F(x,y)=axy+x-x^2-y^2$, with constant Hessian matrixIts eigenvalues are $-2+a$ and $-2-a$. HenceFor $|a|<2$, both eigenvalues are negative and the quadratic polynomial is strictly concave, so the maximum is global on the entire constraint surface. For $|a|>2$, the opposite signs give increasing and decreasing directions. At $a=\pm 2$ there is no stationary point, as already found, rather than a degenerate extremum to classify. No parameter value produces a constrained local minimum.

The Hessian matrix isAt the origin it is the identity, so the origin is a strict local minimum. At $(0,1)$ it is $\mathrm{diag}(3,-1)$, so this point is a saddle point. At either remaining stationary point,which again gives a saddle point. All three saddle points have eigenvalues $3$ and $-1$; there are no local maxima.

The Hessian matrix isIts eigenvalues have opposite signs, so the stationary point is a saddle point. More directly, putting $x=1+u$ and $y=-1+v$ gives $f=u^2+uv-v^2$: it increases along $v=0$ and decreases along $u=0$.

The Hessian matrix has entriesAt each stationary point the mixed entry vanishes, so the signs of the diagonal entries directly classify the local quadratic change:In particular the two local minima are also global minima: the squared term is nonnegative and $\ln (3+y^2)\ge \ln 3$, with equality in both bounds exactly at $(\pm 1,0)$. The local maximum at the origin is not a global maximum, since $f$ grows without bound as $x^2+y^2\to \infty$. The opposite diagonal signs at the other two points establish their saddle point character.

The level set $f=c$ can be rearranged, without changing its domain, asThe requested special contours are thereforeThe first is a parabola, the second a single point, and the last two intersecting straight lines. Moreoverwith equality only at $(0,1)$. Hence $(0,1)$ is a strict global maximum. At $(0,-1)$ the crossing critical contour separates sectors with values above and below $-1$, so it is a saddle point of a scalar function.

As a Hessian matrix check, $H_f(0,1)=\mathrm{diag}(-3,-1)$ is negative definite, while $H_f(0,-1)=\mathrm{diag}(-1,1)$ is indefinite. There is no stationary minimum.

The determinant of the Hessian matrix is $4\cdot 2-3^2=-1<0$. Hence it has one positive and one negative eigenvalue (explicitly $3\pm \sqrt{10}$), so its quadratic form is indefinite. The cubic terms are of higher order and cannot remove these opposite local directions of change. ThereforeFor example the quadratic term is positive along $y=0$ but negative along $y=-3x/2$, confirming the saddle point of a scalar function classification.

The second partial derivatives are $f_{xx}=4$, $f_{xy}=f_{yx}=3+2y$ and $f_{yy}=2+2x+30y$. Evaluating at the origin gives the Hessian matrix

For a stationary point, the two factors from part (a) must vanish. If $y=0$, the second equation gives $x(x^2-1)=0$; if $x=0$, the first gives $y(y^2-1)=0$. Thus the axis points are $(0,0),(\pm 1,0),(0,\pm 1)$.

If both coordinates are nonzero, solve $3x^2+y^2=1$ and $x^2+3y^2=1$, giving $x^2=y^2=1/4$. These four points complete the list of nine stationary points.

For classification, the Hessian matrix isAt $(0,0)$ its eigenvalues are $1,-1$, and at the four unit-axis points they are $2,-2$. These five points are saddle points of a scalar function, all with value zero.

At $(1/2,1/2)$ and $(-1/2,-1/2)$ the eigenvalues are one and two, so both are strict local minima of value $-1/8$. At $(1/2,-1/2)$ and $(-1/2,1/2)$ they are minus one and minus two, so both are strict local maxima of value $1/8$. This follows from the second-derivative test. These extrema are local: the polynomial is unbounded above and below on the full plane.

Near each local minimum, the gradient points outward; near each local maximum, it points inward. The five remaining points have saddle directions, determined by their indefinite Hessian matrices.

On the $x$-axis the exact gradient is $(0,x(x^2-1){)}^T$, and on the $y$-axis it is $(y(y^2-1),0{)}^T$. Thus arrows are perpendicular to the axes and reverse direction at the origin and the unit-axis stationary points.

On the unit circle, substitute $x^2+y^2=1$ into the gradient to obtainIt is radial outward in quadrants I and III, radial inward in quadrants II and IV, and zero at the four axis intersections. These arrows, together with the nearby saddle and extremum patterns, are shown in the right panel of the contour figure. The gradient is normal to every regular level curve and points toward increasing values of $f$.

The Hessian matrix is diagonal:Its eigenvalues are $c$ and $d$. Hence the second-derivative test gives a strict local minimum if $c,d>0$, a strict local maximum if $c,d<0$, and a saddle point of a scalar function if $cd<0$. No degenerate case occurs because both are nonzero.

The local minimum is the unique global minimum, since both separated summands are strictly convex functions and tend to $+\infty$ at either end of the real line. Similarly, when both signs are negative there is a unique global maximum.

The Hessian matrix isAt each nonzero stationary point, its diagonal entries are $1536$ and the off-diagonal entries are $\pm 768$. Its eigenvalues are $768$ and $2304$, both positive. Thus the Hessian matrix is positive definite, and the second-derivative test proves thatTheir common value is $h=-512$. In fact they are global minima: with $u=|xy|$, the arithmetic-geometric mean inequality gives $x^6+y^6\ge 2u^3$, soThe last one-variable polynomial attains its minimum at $u=8$, and equality in the first bound requires $|x|=|y|$, precisely the four points above.