= Solution
{c}
The <gradient> of the given <polynomial> is
$$
\nabla f=(2x+y^2+1,\ 2y(1+x)).
$$
Its second component vanishes when $y=0$ or $x=-1$. The first component then gives exactly three <critical points>:
$$
(-\tfrac12,0),\qquad (-1,1),\qquad (-1,-1).
$$
The <Hessian matrix> and its <determinant> are
$$
H_f=\begin{pmatrix}2&2y\\2y&2+2x\end{pmatrix},\qquad \det H_f=4(1+x-y^2).
$$
At $(-1/2,0)$ the <Hessian matrix> is $\operatorname{diag}(2,1)$, whose <positive definiteness> proves a strict <local minimum> by the <second-derivative test>. At $(-1,\pm1)$ its <determinant> is $-4$, so both points are <saddle points of a scalar function>. Their values and classifications are
$$
\boxed{\begin{aligned}
f(-\tfrac12,0)&=-\tfrac14&&\text{strict local minimum},\\
f(-1,\pm1)&=0&&\text{saddle points}.
\end{aligned}}
$$
There are \b[no <local maxima>].
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