<s>[INST] Stationary points of a multivariable function This question might just be a quick one but I'm slightly confused by the answer I've been provided for this question.
I have the function: $f(x,y) = (x^2+2y^2)e^{-y^2 - x^2}$
I found the partial derivatives:
$f_x = (2x-2x(x^2+2y^2))e^{-y^2-x^2}$
$f_y = (4y-2y(x^2+2y^2))e^{-y^2-x^2}$
I know that the stationary points can be found where $f_x=0$ and/or where $f_y=0$, so that's what I did, I set both $f_x$ and $f_y$ to equal $0$:
\begin{align}
\tag{1}
(2x-2x(x^2+2y^2))e^{-y^2-x^2}&=0\\
\tag{2}
(4y-2y(x^2+2y^2))e^{-y^2-x^2}&=0
\end{align}
I then divided both $(1)$ and $(2)$ by $e^{-y^2-x^2}$ and simplified both equations, this is what I was left with:
\begin{align}
\tag{1'}
x(1-x^2-2y^2)=0\\
\tag{2'}
y(2-x^2-2y^2)=0
\end{align}
This is where I was slightly confused. On the answer sheet I'm provided, the only stationary points they find are $(0,0)$, $(1,0)$ and $(-1,0)$
I understand how $(0,0)$ was found but I'm not sure how $(\pm1,0)$ was found. Did they make the two equations equal to each other then solve?
 [/INST] Let us solve your last system systematically. In $(1')$ there are two possibilities, $x = 0$ or $x^2 = 1-2y^2$.
In the former case, substituting $x = 0$ in $(2')$ yields $y(2-2y^2) = 0$, so you get $y = 0$, $y = 1$ and $y = -1$.
In the latter case, substituting $x^2 = 1-2y^2$ in $(2')$ yields $y = 0$. Now substituting $y = 0$ in $(1')$ yields $x(1-x^2) = 0$, so you get $x = 0$, $x = 1$ and $x = -1$.
All in all, you have five critical points: $(0,0)$, $(0,\pm 1$), $(\pm 1,0)$.
Wolfram Alpha gives you a nice diagram of the critical points:

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