Solution
= Solution
The <gradient> components are
$$
u_x=x^2(3-2x^2)e^{-x^2-y^2},\qquad u_y=-2yx^3e^{-x^2-y^2}.
$$
Away from the vertical axis, $x\ne0$, so $u_y=0$ forces $y=0$. The other <stationary point> equation then gives $x^2=3/2$. Therefore the two requested points are
$$
\boxed{\left(\sqrt{\frac32},0\right),\qquad\left(-\sqrt{\frac32},0\right).}
$$
Every point with $x=0$ also has zero <gradient>, but those points are explicitly excluded.