= Solution
Put $G=(x^2+y^2)/2+ax^4/4$, so $f=e^G>0$. Since $f_x=fG_x$ and $f_y=fG_y$, its <stationary points> satisfy
$$
y=0,\qquad x(1+ax^2)=0.
$$
The origin occurs for every real $a$. There are two additional points only for $a<0$, at $x=\pm1/\sqrt{-a}$.
At a <stationary point>, the <Hessian matrix> simplifies because $\nabla G=0$:
$$
H_f=fH_G=f\begin{pmatrix}1+3ax^2&0\\0&1\end{pmatrix}.
$$
At the origin this is $I$, so \b[$(0,0)$ is a strict local minimum for every $a$], with $f=1$. If $a\geq0$ it is also the unique global minimum, since $G\geq0$ with equality only there. For $a<0$, the two extra points have $1+3ax^2=-2$, so their <Hessian> has opposite-sign <eigenvalues>. Thus
$$
\boxed{a<0:\ \left(\pm\frac1{\sqrt{-a}},0\right)\text{ are saddle points},\qquad
f=\exp\!\left(-\frac1{4a}\right).}
$$
There are no <local maxima>. When $a<0$, the origin is not a global minimum: $f(x,0)\to0$ as $|x|\to\infty$, but $f$ never attains zero.
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