Critical Points of xy e^(-x^2-y^2)

Factor first, then enumerate the cases

Each partial needs the product rule, and the two resulting terms share a common factor that is worth pulling out immediately.

The exponential is never zero, so it divides out of both equations and leaves two factored conditions.

Now enumerate. The first equation gives $y = 0$ or $x = \pm\frac{1}{\sqrt{2}}$. The second gives $x = 0$ or $y = \pm\frac{1}{\sqrt{2}}$. Pairing them off:

• $y = 0$ with $x = 0$ gives the origin.
• $y = 0$ with $y = \pm\frac{1}{\sqrt{2}}$ is a contradiction, so nothing.
• $x = \pm\frac{1}{\sqrt{2}}$ with $x = 0$ is a contradiction, so nothing.
• $x = \pm\frac{1}{\sqrt{2}}$ with $y = \pm\frac{1}{\sqrt{2}}$ gives four points, one in each quadrant.

Five points in total, and because the enumeration was exhaustive, that is the complete list.

Classifying all five at once

The second partials are more pleasant than they look, especially the mixed one, which factors completely.

At the origin the first two vanish and $f_{xy} = 1$, so $D = -1$ and the point is a saddle. Near $(0,0)$ the exponential is close to 1 and $f$ behaves like $xy$, positive in two opposite quadrants and negative in the other two.

At the four outer points, $x^{2} = y^{2} = \frac{1}{2}$, so both factors in $f_{xy}$ are zero and the mixed partial disappears. The exponent is $-1$ there, and $3 - 2x^{2} = 2$, so the diagonal entries reduce to $f_{xx} = f_{yy} = -4xy\,e^{-1}$.

The discriminant is the same positive number at all four, so $f_{xx}$ alone decides. Where $x$ and $y$ agree in sign, $xy = \frac{1}{2}$ and $f_{xx} = -\frac{2}{e} < 0$: local maxima with $f = \frac{1}{2e} \approx 0.184$. Where they differ in sign, $xy = -\frac{1}{2}$, $f_{xx} = \frac{2}{e} > 0$, and those are local minima with $f = -\frac{1}{2e}$.

Those extremes are global. Since $|xy| \le \frac{x^{2}+y^{2}}{2}$, writing $r^{2} = x^{2}+y^{2}$ gives $|f| \le \frac{r^{2}}{2}e^{-r^{2}}$, and the right side is largest at $r^{2} = 1$, where it equals $\frac{1}{2e}$.

The mistake students make

The most common loss is the origin itself. Students factor the two equations, jump to the interesting case $x = \pm\frac{1}{\sqrt{2}}$ and $y = \pm\frac{1}{\sqrt{2}}$, report four critical points and stop. The case $x = y = 0$ satisfies both equations just as legitimately, and it is the only saddle on the surface. Enumerating all four pairings on paper, including the ones that die, is what prevents this.

The second error is dropping a term from the product rule. Differentiating $xy\,e^{-x^{2}-y^{2}}$ in $x$ means differentiating a product of $xy$ and the exponential, so both pieces contribute.

Keep only the second term and you get $-2x^{2}y\,e^{-x^{2}-y^{2}}$, with $-2xy^{2}e^{-x^{2}-y^{2}}$ from the same slip in $y$. Both vanish on the whole $x$-axis and the whole $y$-axis, so the answer becomes two full lines of critical points in place of the single saddle at the origin, and the $1 - 2x^{2}$ factor that locates the four real extrema never appears.

A cheap check: the four outer critical points must sit at $x^{2} = y^{2} = \frac{1}{2}$, distance 1 from the origin. If your answer puts them anywhere else, or puts them on an axis, the product rule is the first place to look.