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Put $S={\theta }_1+{\theta }_2$ and $D={\theta }_1-{\theta }_2$. The gradient has components $3\sin S+\sin D$ and $3\sin S-\sin D$, so a stationary point requires $\sin S=\sin D=0$. In the closed square, $S\in [0,2\pi ]$ and $D\in [-\pi ,\pi ]$, and the compatible possibilities are precisely the four corners and $(\pi /2,\pi /2)$.

The Hessian matrix isIts eigenvectors $(1,1)$ and $(1,-1)$ have eigenvalues $6\cos S$ and $2\cos D$. The second-derivative test gives the following classification:

The corner classifications also hold for the smooth extension beyond the square, and hence for the restricted region. The global bounds $-4\le U\le 4$ show that the stated minima and maxima are global within it. The central point has opposite curvature along the two diagonal directions.

The gradient is $(f_x,f_y)=(2-2xy,1-x^2)$. Both partial derivatives vanish at $(1,1)$, so it is a stationary point.

The second component of the gradient forces $x=\pm 1$. The first then requires $xy=1$. Apart from $(1,1)$, the only stationary point is

Set $g(x,y)=y-(x-2{)}^2$. The gradient of $g$ never vanishes, so the Lagrange multiplier equations $\nabla f=\lambda \nabla g$ apply at a constrained stationary point. They are $4x=-2\lambda (x-2)$, $2y=\lambda$, and $y=(x-2{)}^2$. Eliminating $\lambda ,y$ givesWith $t=x-2$, this becomes $t^3+t+2=(t+1)(t^2-t+2)=0$. The quadratic factor is $(t-1/2{)}^2+7/4>0$, so the only real possibility is $t=-1$. ThereforeThe level curve through this point is the ellipse $2x^2+y^2=3$, meeting the coordinate axes at $(\pm \sqrt{3/2},0)$ and $(0,\pm \sqrt{3})$. Its tangent line at $(1,1)$ has slope $-2x/y=-2$, equal to the derivative $2(x-2)$ of the constraint parabola there. The two curves are tangent.

The parabola touches this ellipse from outside; every other point on the constraint lies on a larger contour. An exact check of this geometric argument iswith equality only at $x=1$. Thus the point is the unique constrained global minimum.

Inside the triangle, both $x$ and $y$ are nonzero. Thus $\nabla f=0$ reduces to $1-2x+y=0$ and $1-x+2y=0$, whose unique solution is $(1/3,-1/3)$. At that pointThe Hessian matrix has positive eigenvalues $1$ and $1/3$, so this is a strict local minimum. It is also the minimum on the closed triangle: $f=0$ on the boundary, $f<0$ inside, and there is no other interior stationary point. Therefore

The gradient is $(-y^2-2x-3x^2,2y(1-x))$. Its second component vanishes only if $y=0$ or $x=1$. The latter gives the impossible first-component equation $-y^2-5=0$. With $y=0$, the first component gives $x=0$ or $x=-2/3$. The Hessian matrix isAt $(0,0)$ it has eigenvalues $-2,2$, giving a saddle point. At $(-2/3,0)$ its eigenvalues are $2,10/3$, giving a strict local minimum, of value $-4/27$. These are all real stationary points. The minimum is local: for example $g(x,0)\to -\infty$ as $x\to +\infty$.

The denominator $D=x^2+xy+y^2=(x+y/2{)}^2+3y^2/4$ is positive away from the origin. The partial derivatives areThey both vanish exactly onThese are lines of stationary points, with values $1/3$ and $-1$, respectively.

Put $X=x-1$ and $H=e^{-(X^2+y^2)/2}>0$. The gradient components are $z_x=y(1-X^2)H$ and $z_y=X(1-y^2)H$. If either $X$ or $y$ vanishes, both must vanish. Otherwise both squared coordinates must be one. Therefore the five stationary points in the original $(x,y)$ coordinates are

As printed, the function is undefined at the origin. For $r^2=x^2+y^2>0$, its gradient isThe scalar prefactor is positive, and $(x,y)\ne (0,0)$, so this gradient never vanishes. Thus there are no stationary points on the stated domain. Its infimum is three, approached as $r\to 0$, but that value is not attained there.

If the intended definition includes the natural extension $f(0,0)=3$, that extension has a unique strict global minimum at the origin, with value three. Indeed, $e^{-1/r^2}>0$ off zero, and $e^{-1/r^2}/r\to 0$ proves differentiability at zero with zero gradient. More strongly, subtracting three gives the radial Gaussian flat function: every derivative of that difference vanishes at the origin. The vanishing Hessian matrix does not turn this minimum into an inflection point. These are two different domain interpretations; the extension must be stated rather than silently inserted.

Write $h(x,y)=y(r-4)$, where $r=\sqrt{x^2+y^2}$. Because this hyperbolic function has strictly positive derivative $\cosh$, stationary points under an increasing outer function have the same positions and types for $g=\sinh h$ and $h$. For $r>0$,The first equation requires $x=0$ or $y=0$. If $y=0$, the second gives $|x|=4$. If $x=0$, it gives $2|y|=4$. These are all the nonzero stationary points. At the origin, $h=-4y+O(r^2)$, so its gradient is $(0,-4)$; the origin is not stationary.

At $(0,2)$ the Hessian matrix of $h$ is $\mathrm{diag}(1,2)$, and at $(0,-2)$ it is $\mathrm{diag}(-1,-2)$. At $(4,0)$ and $(-4,0)$ its Hessian matrices are respectively $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and $\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)$. ThereforeThe extrema are local: $g$ is unbounded above and below on the full plane.

For the level curve geometry, a level $g=c$ is $y(r-4)=\mathrm{arsinh}c$. In particular, the zero level curve is the union of the entire $x$ axis and the circle $r=4$, crossing at the two saddle points. Inside the circle, $g<0$ above the axis and $g>0$ below it; outside, these signs reverse. Levels between $-\sinh 4$ and zero have closed loops around the upper minimum as well as unbounded lower-half-plane components. Reflection across the $x$ axis reverses the sign and gives the loops about the lower maximum. At the two extremal heights, the corresponding closed loop shrinks to its extremum. These features determine the sketch.

The first partial derivatives factor asBoth vanish at the origin, verifying that it is a stationary point. Since $2x^2+1>0$, every stationary point has $y=0$; the first equation then gives $x=0,\pm 1$. Thus one requested additional point is $\boxed{(1,0)}$; the other is $\boxed{(-1,0)}$.

Differentiating the hyperbolic cosine with the chain rule gives $g^{\prime }(x)=3{\cosh }^{2}x\sinh x$. For real $x$, $\cosh x>0$, so the only zero of the derivative is $\sinh x=0$, namely $x=0$. ThusSince $\cosh x\ge 1$, this stationary point is the global minimum.

Since $f>0$, a level $f=c$ can occur only for $c>0$. Taking its natural logarithm gives $xy=\ln c$. Thus $c>1$ produces rectangular hyperbolas in the first and third quadrants, $0<c<1$ produces them in the second and fourth quadrants, and $c=1$ consists of both coordinate axes.

The gradient $e^{xy}(y,x)$ vanishes only at $(0,0)$, where $f=1$. The Hessian matrix there is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, with negative determinant. Equivalently, $f(x,x)>1$ while $f(x,-x)<1$ for nonzero small $x$. Hence the only stationary point is a saddle point at $(0,0,1)$ on the graph; its location in the contour plane is the origin.

The quotient rule simplifies the derivative toIts zeros are $-5$ and $-1$, both in the domain. Substitution gives the two stationary pointsThe first is a local maximum and the second a local minimum, since the numerator of $y^{\prime }$ changes sign from positive to negative at $-5$ and from negative to positive at $-1$. The denominator of the original rational function vanishes atThe numerator is nonzero at both values, so these discontinuities are poles, not removable holes.

The product rule gives the two partial derivativesThey both vanish at the origin. The exponential function is never zero, so any further stationary point must satisfy $x=0$ or $x^2-y^2=2$, and $y=0$ or $y^2-x^2=2$. Having both coordinates nonzero would require both opposite differences to be $2$, which is impossible. If $y=0$, the nonzero choices are $x=\pm \sqrt{2}$; if $x=0$, they are $y=\pm \sqrt{2}$. Hence the full list is

Minimize the squared distance $F=x^2+y^2$, which has the same minimizers as the distance. The Lagrange multiplier equations for the constraint $G=x^2+xy-1=0$ areThe constraint gradient cannot vanish on this branch, since its second component is $x>0$. Eliminate $\lambda =2y/x$ to obtain $x^2=2xy+y^2$. Put $t=y/x$; then $t^2+2t-1=0$, so $t=-1\pm \sqrt{2}$. But the constraint says $x^2(1+t)=1$, requiring $1+t>0$ and selecting $t=\sqrt{2}-1$. It follows thatTo prove a global minimum on the whole $x>0$ branch, substitute $y=1/x-x$ into $F$:The arithmetic-geometric mean inequality proves this bound, with equality only when $2x^2=x^{-2}$, namely $x=2^{-1/4}$. Thus the stationary point is the unique closest point on the required branch. Numerically it is approximately $(0.840896,0.348311)$ and its distance is $0.910180$ in coordinate units.

The derivative is $f^{\prime }(x)=2x\cos (x^2-2)$. Thus a stationary point occurs either at $x=0$ or where $x^2-2=\pi /2+k\pi$. Since $-2\le x^2-2<2$ on the specified open interval, the only allowed cosine zeros are $-\pi /2$ and $\pi /2$. The five stationary abscissae areTheir ordinates are $-\sin 2$, $-1$, and $1$, respectively. For completeness, $f^{\prime \prime }=2\cos (x^2-2)-4x^2\sin (x^2-2)$ shows that the inner pair are local minima and the outer pair are local maxima; $x=0$ is also a local maximum because $2\cos 2<0$.

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.