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The two coefficients are the corresponding partial derivatives of $e^{x+y}$. Therefore the differential form is exact, withThe arbitrary constant does not change the total differential.

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.

Set $T=t+a>0$. The similarity solution uses $y=(x+b)T^{-1/2}$, whose partial derivatives at fixed $x$ or $t$ areApply the multivariable chain rule to $u=T^{-1/2}v(y)$:Substituting in the heat equation and cancelling the positive common factor $T^{-3/2}$ gives

Taking each partial derivative of the scalar potential givesThe factor $5y^4$ in the middle component comes from the chain rule. The potential is smooth, so the curl of a gradient identity gives $\boxed{\nabla \times \mathbf{F}=0}$.

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 $|t|\to \infty$.

Write $\mathbf{a}=(a_x,a_y,a_z)$. The cross product gives ${\mathbf{v}}_1=(a_{y}z-a_{z}y,a_{z}x-a_{x}z,a_{x}y-a_{y}x)$, whose three diagonal partial derivatives vanish. For the second vector field, direct differentiation gives ${\partial }_x(2xy)=2y$, ${\partial }_y(x^2+3y^2+z^2)=6y$, and ${\partial }_z(2yz)=2y$. Thus

Taking the second partial derivatives of the polynomial givesAdding cancels the pure fourth powers and leavesThus direct substitution verifies the stated Poisson equation.

The potential is smooth everywhere, so its mixed partial derivatives commute. The curl of a gradient therefore vanishes:

Write $A=f_{xx}$, $B=f_{xy}$, $C=f_{yy}$ at the stationary point, and let $D=AC-B^2\ne 0$. Under the usual second-order Taylor expansion hypothesis, the second-derivative test givesIndeed the quadratic term is $(A{\xi }^2+2B\xi \eta +C{\eta }^2)/2$. For $A\ne 0$, completing its square gives $[A(\xi +B\eta /A{)}^2+(D/A){\eta }^2]/2$. The two coefficients have the same positive sign, the same negative sign, or opposite signs in the respective cases. A negative $D$ also forces the two Hessian eigenvalues to have opposite signs, including the case $A=0$.

There is a qualification to the printed assumptions: existence of second partial derivatives only at the point does not ensure the needed quadratic approximation. Continuous second partial derivatives in a neighborhood are sufficient. The existence of second partial derivatives does not justify the Hessian test: for example, the function $g=x^2+y^2-6x^2{y}^2/(x^2+y^2)$ away from the origin, extended by $g(0,0)=0$, has Hessian $2I$ at the origin but $g(t,t)=-t^2$ and $g(t,0)=t^2$. Thus the classification above uses the standard additional regularity assumption, which the smooth function in part (b) satisfies.

For a twice continuously differentiable scalar potential, the curl of its gradient isEquality of mixed partial derivatives makes every component zero. ThusThis is the curl of a gradient identity.

The original PDF has $8y^3-4x$ in the denominator and $dy/dx$ on the left. Rearrange it into the differential formThe cross partial derivatives of its coefficients are both four, so it is an exact differential form. Integrating its first coefficient in $x$ gives a potential $F=\frac{5}{2}{x}^2+4xy+h(y)$. Matching $F_y=4x-8y^3$ gives $h(y)=-2y^4$. Therefore an implicit general solution isDifferentiating verifies $y^{\prime }=(5x+4y)/(8y^3-4x)$. A solution graph is understood only on intervals where its denominator is nonzero; the printed quotient does not define a derivative when $8y^3=4x$.

Differentiate the angular relation locally, or use $d\varphi =(xdy-ydx)/(x^2+y^2)$. Holding each other coordinate fixed givesThese are local partial derivatives on a consistent angular branch, independent of where its branch cut is placed.

Apply the multivariable chain rule to $f(x,y)=g(r(x,y),\varphi (x,y))$. Using the coordinate partial derivatives above givesEquivalently, ${\partial }_x=\cos \varphi {\partial }_r-(\sin \varphi /r){\partial }_{\varphi }$ and ${\partial }_y=\sin \varphi {\partial }_r+(\cos \varphi /r){\partial }_{\varphi }$. Their coefficients vary with position and must also be differentiated when applying them twice.

Write $c=\cos \varphi$ and $s=\sin \varphi$. The needed coefficient derivatives are $c_x=s^2/r$, $s_x=-sc/r$, $c_y=-sc/r$ and $s_y=c^2/r$. Apply the product rule to both terms of $f_x=c{g}_r-(s/r)g_{\varphi }$ and use the chain rule for the derivatives of $g$. Collecting terms givesSimilarly, differentiating $f_y=s{g}_r+(c/r)g_{\varphi }$ givesOn addition, both the mixed partial derivative terms and the first angular derivative terms cancel. Using $s^2+c^2=1$, the Laplacian in polar coordinates is

The curl of the constant vector field $\mathbf{a}\times \mathbf{b}$ vanishes. The curl of the position field $\mathbf{r}=(x,y,z)$ also vanishes, since all its cross-coordinate partial derivatives are zero. Thus

On a smooth implicit branch, put $Q=x+z^2$ and require $Q\ne 0$. This is exactly the implicit function theorem condition, since the defining equation has vertical partial derivative $3Q$. Differentiating the cubic relation in $x$ and $y$ givesFor a second derivative, the dependence of $Q$ on $z$ must also be included: $Q_x=1+2z{z}_x$ and $Q_y=2z{z}_y$. ThusThe required cancellation follows immediately:The verification is local to smooth branches with $x+z^2\ne 0$; the implicit relation does not guarantee a differentiable function at its singular points.

For $r>0$, the partial derivatives of radius are $r_x=x/r$, $r_y=y/r$ and $r_z=z/r$. Applying the chain rule to the outer sine and then the radial profile gives the Cartesian coordinates of the gradient:At the origin these coordinate formulas require a differentiable radial extension; differentiability of $f$ as a one-variable profile alone does not guarantee it.

The vector field has only a vertical component, namely $f(r)$. Its divergence is therefore its vertical partial derivative:

Let $M=2x+e^y$ and $N=x{e}^y-\cos y$. Their cross partial derivatives agree: $M_y=e^y=N_x$. On the simply connected plane, this makes the differential form exact. Integrating $F_x=M$ gives $F=x^2+x{e}^y+h(y)$; matching $F_y=N$ then requires $h^{\prime }(y)=-\cos y$, so one potential isThe ordinary differential equation states $dF/dx=0$ along a solution. Its initial condition fixes the constant to $F(1,\pi /2)=e^{\pi /2}$. Hence the solution is specified implicitly byAt the initial point $F_y=e^{\pi /2}\ne 0$, so the implicit function theorem guarantees a unique local branch $y(x)$ through that point.

Work on an open domain with $xy>0$ and $x\ne 0$, where the displayed square root is real and differentiable. Split $u=v+h$ with $v=\varphi (xy)$ and $h=\sqrt{xy}\psi (y/x)$. For the first term, the chain rule givesso $x^2{v}_{xx}-y^2{v}_{yy}=0$.

For the second term, set $t=y/x$. Its first partial derivatives areThey show $x{h}_x+y{h}_y=h$, the degree-one case of Euler theorem for homogeneous functions. Differentiating that identity with respect to each variable yieldsThe mixed partial derivatives agree here by the chain rule. Multiplying these equations by $x$ and $y$, respectively, proves the second derivative identity for a degree-one homogeneous function:Adding the results for $v$ and $h$ verifies the partial differential equationEquivalently, $h(\lambda x,\lambda y)=\lambda h(x,y)$ for positive $\lambda$, so the same proof follows directly from its homogeneous function property. The statement is local to a domain where the given expression is defined; it does not require assigning values on $x=0$.