Finding extremal points on $f(x,y)$ This is the equation: 
$$f(x,y) = xye^\left(-\frac{1}{2}(x^2 + y^2)\right)$$
This is what I've done:
$$\nabla f(x,y) = \begin{bmatrix}
(1-x^2)ye^\left(-\frac{1}{2}(x^2 + y^2)\right) \\ 
(1-y^2)xe^\left(-\frac{1}{2}(x^2 + y^2)\right)
\end{bmatrix}$$
Here's the thing I'm worried about, to find when $\nabla f(x,y) = 0$, i set each equation $= 0$.
$$\begin{align}
(1-x^2)ye^\left(-\frac{1}{2}(x^2 + y^2)\right) &= 0 \\
y e^\left(-\frac{1}{2}(x^2 + y^2)\right) &= x^2ye^\left(-\frac{1}{2}(x^2 + y^2)\right) \\
1 &= x^2 \\
x &= \pm 1\end{align}
$$
Is this legal, or do i lose some solutions when I divide away everything?
 A: Dividing by $y$ is one of the problematic steps.
Just start with the partial derivative, and divide by the exponential term (which is never zero) to get $(1-x^2)y=0$. From here you see either $1-x^2=0$ or $y=0$, from which you get $x=\pm 1$ or $y=0$.
If you do this for the other partial derivative you get $y \pm 1$ or $x=0$.
Combine this with the above to gather all the critical points.
A: For another approach, to avoid slogging through the partial derivates, we can exploit the symmetry of $f$. If we switch to polar coordinates, we have 
$f(r,t) = xye^\left(-\frac{1}{2}(x^2 + y^2)\right)=\frac{r^{2}}{2}\sin 2t\cdot e^\left (-\frac{1}{2}(r^2)\right).$
Inspection shows that $f$ has extrema at $t=\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}\ $ and $\frac{7\pi}{4},\ $  and in each case $f(r,t)=\pm \frac{r^{2}}{2}e^\left (-\frac{1}{2}(r^2)\right).$ Now it's an easy Calc I exercise to show that $f$ has a local maximum at $r=\sqrt 2,t=\frac{\pi}{4},\ $ or $x=1,y=1,\ $ and using the symmetry of $f$ again, we can read off the other local max, and the two local mins. 
