Solution
= Solution
Put $X=x-1$ and $H=e^{-(X^2+y^2)/2}>0$. The <gradient> components are $z_x=y(1-X^2)H$ and $z_y=X(1-y^2)H$. If either $X$ or $y$ vanishes, both must vanish. Otherwise both squared coordinates must be one. Therefore the five <stationary points> in the original $(x,y)$ coordinates are
$$
\boxed{(1,0),\quad(0,1),\quad(0,-1),\quad(2,1),\quad(2,-1).}
$$