= Solution
{c}
The <critical points> satisfy
$$
f_x=2x+y=0,\qquad f_y=3y^2+x=0.
$$
Substitute $x=-y/2$ to obtain $y(3y-1/2)=0$. Thus the only <critical points> are $(0,0)$ and $(-1/12,1/6)$.
The <Hessian matrix> and <Hessian determinant> are
$$
H=\begin{pmatrix}2&1\\1&6y\end{pmatrix},\qquad D=12y-1.
$$
At $(0,0)$, $D=-1$, so this is a <saddle point>. At $(-1/12,1/6)$, $D=1$ and $f_{xx}=2>0$, so this is a strict <local minimum>. Therefore
$$
\boxed{(0,0):\text{ saddle};\qquad(-1/12,1/6):\text{ strict local minimum}.}
$$
The value at the <local minimum> is $-1/432$; there is no <local maximum>.
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