= Solution
The stationarity equations are $4x^3-16y=0$ and $4y^3-16x=0$. Thus $y=x^3/4$ and $x=y^3/4$, so $x(x^8-256)=0$. The real solutions are $(0,0)$, $(2,2)$ and $(-2,-2)$. The <Hessian matrix> is
$$
H=\begin{pmatrix}12x^2&-16\\-16&12y^2\end{pmatrix}.
$$
At the origin its determinant is $-256$, giving a saddle. At either nonzero stationary point it has eigenvalues $32,64$, so both are strict minima, with value $-32$:
$$
\boxed{(0,0):\text{saddle},\quad f=0;\qquad (2,2),(-2,-2):\text{minima},\quad f=-32.}
$$
They are also global minima, because $f=(x^2-y^2)^2+2(xy-4)^2-32\geq-32$, with equality exactly at those two points. This is the special coefficient choice in a <quartic double well with bilinear coupling>.
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