Second partials of x e^y + y e^x

Term by term, holding the other variable still

Differentiate in $x$ with $y$ frozen. In the first term $x e^{y}$, the factor $e^{y}$ is a constant, so the derivative is just $e^{y}$. In the second term $y e^{x}$, the constant is $y$ and $e^{x}$ differentiates to itself.

Now take a second $x$ derivative of $f_x = e^{y} + y e^{x}$. The term $e^{y}$ is a constant in $x$ and dies, leaving only $y e^{x}$.

For the mixed partial, differentiate $f_x = e^{y} + y e^{x}$ with respect to $y$. Now the first term survives as $e^{y}$ and the second contributes $e^{x}$, because $y$ is the variable and $e^{x}$ is the constant.

The mistake: keeping a term that should have vanished

Terms disappear and reappear as you switch variables, and that is where marks are lost. Going from $f_x = e^{y} + y e^{x}$ to $f_{xx}$, the $e^{y}$ must vanish, because it is constant in $x$. Going from the same $f_x$ to $f_{xy}$, that identical $e^{y}$ is the term that survives while nothing else does. Students who carry the same term through both steps get $f_{xx} = e^{y} + y e^{x}$, which is the first partial written out again.

Before each step, ask which factors contain the variable you are differentiating in and cross out the rest. Here no term ever needs the product rule, since in $x e^{y}$ the two factors depend on different variables, and the same is true of $y e^{x}$.

Clairaut symmetry, and the one critical point

Take the other route. Start from $f_y = x e^{y} + e^{x}$ and differentiate in $x$: the first term gives $e^{y}$ and the second gives $e^{x}$, so the answer is $e^{x} + e^{y}$ again.

What makes this example worth remembering is that the surviving terms swap places. Going $x$ first, the $e^{y}$ in the answer comes from the term $x e^{y}$; going $y$ first, it comes from the same term but through the opposite factor. The totals match because Clairaut's theorem says they must, given that both mixed partials are continuous everywhere.

The function is also symmetric under swapping $x$ and $y$, so $f_{xx}(x,y) = y e^{x}$ and $f_{yy}(x,y) = x e^{y}$ are mirror images. The discriminant $D = xy\,e^{x+y} - (e^{x}+e^{y})^{2}$ is worth evaluating only where both first partials vanish. Setting $e^{y} = -y e^{x}$ and $x e^{y} = -e^{x}$, then substituting the first into the second, gives $xy = 1$, with $y < 0$ forced because $-y e^{x}$ has to come out positive. The one point meeting both conditions is $(-1,-1)$.

So this surface has exactly one critical point, a saddle at $(-1,-1)$ of height $f(-1,-1) = -2/e$. Everywhere else the discriminant tells you nothing, because the second derivative test only classifies points where the gradient is already zero.