= Solution
The stationary equations are
$$
4x^3-6x+2y=0,\qquad 2x+2y=0.
$$
Thus $y=-x$ and $4x(x^2-2)=0$, giving
$$
(0,0),\qquad(\sqrt2,-\sqrt2),\qquad(-\sqrt2,\sqrt2).
$$
The <Hessian matrix> is
$$
H=\begin{pmatrix}12x^2-6&2\\2&2\end{pmatrix}.
$$
At either nonzero stationary point,
$$
H=\begin{pmatrix}18&2\\2&2\end{pmatrix},
\qquad \det H=32>0,
$$
so both are strict local minima.
Along a line through the origin,
$$
g(t)=f(t\cos\gamma,t\sin\gamma)
=t^2q(\gamma)+t^4\cos^4\gamma,
$$
where
$$
q(\gamma)=-3\cos^2\gamma
+2\sin\gamma\cos\gamma+\sin^2\gamma.
$$
When $\cos\gamma\ne0$ and $r=\tan\gamma$,
$$
q(\gamma)=\cos^2\gamma(r-1)(r+3).
$$
Consequently the origin is a local minimum along the line when
$$
\boxed{\tan\gamma\leq-3\quad\text{or}\quad\tan\gamma\geq1},
$$
including the equality cases because the positive quartic term then leads. It is also a minimum on the vertical line. It is a local maximum along the line when
$$
\boxed{-3<\tan\gamma<1}.
$$
In the first case the graph is locally bowl-shaped. In the second it initially bends downward from the origin, but the positive quartic term eventually turns it upward, producing the usual double-well profile.
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