I'm trying to solve the following system:
$$ \begin{cases} e^{-x^2-y^2}-2xe^{-x^2-y^2}(x+y)=0\\ e^{-x^2-y^2}-2ye^{-x^2-y^2}(x+y)=0\\ \end{cases} $$
for both $x$ and $y$.
The problem is that from the first I can get $y=\frac{1-2x^2}{2x}$ and plugging this back says that $x \in \mathbb{R} \setminus \{0\}$. Also, wouldn't plugging $x$ back into $y$ give yet that $y \in \mathbb{R} \setminus \{0\}$?
I haven't tried the second one, but the first one seems to produce such confusing results that I'm not sure whether I'm attempting to solve the system in the correct way.
The second equation seems to give
$$y=\pm \frac{\sqrt{x^2+2}}{2}-\frac{x}{2}$$
How does this give me any way to find out any other solution than $x,y \not = 0$?
\begin{cases} e^{-x^2-y^2}-2xe^{-x^2-y^2}(x+y)=0\\ e^{-x^2-y^2}-2ye^{-x^2-y^2}(x+y)=0\\ \end{cases}
Dividing both equations by the positive number $e^{-x^2-y^2}$ yields
\begin{cases} 1-2x(x+y)=0\\ 1-2y(x+y)=0\\ \end{cases}
which can be rewritten
\begin{cases} 2x(x+y)=1\\ 2y(x+y)=1\\ \end{cases}
Adding and subtracting them gives
\begin{cases} (x+y)^2=1\\ (x+y)(x-y)=0\\ \end{cases}
If $x+y=0$ the first cannot be satisfied, so we must have $x=y$, which when we plug into the first we get
$$x=y=\pm \frac 12$$
Thank you to John Wayland Bales for the first part of this.