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?

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.
