= Solution
The <critical points> satisfy both first <partial derivatives> equal to zero:
$$
h_x=2x+y=0,\qquad h_y=x+3y^2=0.
$$
Substitute $x=-y/2$ into the second equation to get $y(3y-1/2)=0$. Thus the candidates are $(0,0)$ and $(-1/12,1/6)$.
The <Hessian matrix> and its determinant are
$$
H=\begin{pmatrix}2&1\\1&6y\end{pmatrix},\qquad \det H=12y-1.
$$
At $(0,0)$ the determinant is $-1$, so the <Hessian matrix> is indefinite and the point is a \b[<saddle point>]. At $(-1/12,1/6)$ the determinant is $1$ and $h_{xx}=2>0$, so the <Hessian matrix> is positive definite and the point is a \b[strict <local minimum>], with value
$$
\boxed{h(-1/12,1/6)=-\frac1{432}}.
$$
There is no local maximum. Along $x=0$, $h(0,y)=y^3$ is unbounded above and below, so there are no global extrema either.
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