= Solution
A <critical point> requires $f_x=f_y=0$. Since $y^4+y^2+1>0$ for every real $y$, the first <partial derivative> forces $x=0$. The second then reduces to $3y^2-3=0$, so the only <critical points> are $(0,1)$ and $(0,-1)$.
At these points the <Hessian matrix> is diagonal:
$$
H_f(0,1)=\begin{pmatrix}6&0\\0&6\end{pmatrix},\qquad H_f(0,-1)=\begin{pmatrix}6&0\\0&-6\end{pmatrix}.
$$
The first is a <positive-definite matrix>, giving a strict local minimum; the second is an <indefinite matrix>, giving a <saddle point>. Their values are $f(0,1)=-1$ and $f(0,-1)=3$. Hence
$$
\boxed{(0,1):\text{ strict local minimum, value }-1;\qquad(0,-1):\text{ saddle point, value }3;\qquad\text{no local maxima}.}
$$
The minimum is local: along $x=0$, $f(0,y)=y^3-3y+1$ is unbounded below as $y\to-\infty$.
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