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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?

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    This is were the partial derivative of $x$ is zero. You have to find the same for the $y$ and then you get the extremal points – 2017-02-21
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    Did you see this https://en.wikipedia.org/wiki/Second_partial_derivative_test ? – 2017-02-21
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    But $y = 0$ would also yield zero. – 2017-02-21
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    You can use an "align" environment in the TeX code snippet to make & align stuff. – 2017-02-21

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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.

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    So the points I get are ${(1,1),(1,-1),(-1,1),(-1,-1),(0,0)}$ – 2017-02-21
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    Yes, the $(0,0)$ is also an answer @Viktor. – 2017-02-21
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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.