= Solution
Set both components of the <gradient> to zero. After dividing out the nonzero exponential, they are equivalent to
$$
x(1-4xy)+2y=0,\qquad y(1-4xy)+2x=0.
$$
Adding and subtracting gives $(x+y)(3-4xy)=0$ and $(x-y)(1+4xy)=0$. If $x=y$, the first equation reduces to $x(3-4x^2)=0$, giving the origin and the two diagonal points. If $x=-y$, it reduces to $x(4x^2-1)=0$, giving the origin and the two opposite-diagonal points. The alternative simultaneous values $xy=3/4$ and $xy=-1/4$ are incompatible, so these cases exhaust the possibilities. Thus the five <stationary points> are
$$
\boxed{(0,0),\quad\left(\frac{\sqrt3}{2},\frac{\sqrt3}{2}\right),\quad\left(-\frac{\sqrt3}{2},-\frac{\sqrt3}{2}\right),\quad
\left(\frac12,-\frac12\right),\quad\left(-\frac12,\frac12\right).}
$$
They are marked in both panels of the figure.
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