= Solution
The <product rule> gives the two <partial derivatives>
$$
g_x=x(2-x^2+y^2)e^{-(x^2+y^2)/2},\qquad
g_y=-y(2+x^2-y^2)e^{-(x^2+y^2)/2}.
$$
They both vanish at the origin. The <exponential function> is never zero, so any further <stationary point> must satisfy $x=0$ or $x^2-y^2=2$, and $y=0$ or $y^2-x^2=2$. Having both coordinates nonzero would require both opposite differences to be $2$, which is impossible. If $y=0$, the nonzero choices are $x=\pm\sqrt2$; if $x=0$, they are $y=\pm\sqrt2$. Hence the full list is
$$
\boxed{(0,0),\quad(\pm\sqrt2,0),\quad(0,\pm\sqrt2).}
$$
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