= Solution
{c}
To find a <critical point>, apply <multivariable Newton iteration> to the <gradient>, rather than to the scalar value $F$. The <gradient> and <Hessian matrix> are
$$
\nabla F=\begin{pmatrix}2x+2y\\2x-2y+3y^2\end{pmatrix},\qquad
\nabla^2F=\begin{pmatrix}2&2\\2&-2+6y\end{pmatrix}.
$$
Write the correction as $\mathbf s_0=\mathbf x_1-\mathbf x_0$. At $\mathbf x_0=(-1,-1)$, <Newton's method in optimization> requires
$$
\begin{pmatrix}2&2\\2&-8\end{pmatrix}\mathbf s_0=-\begin{pmatrix}-4\\3\end{pmatrix}=\begin{pmatrix}4\\-3\end{pmatrix}.
$$
Subtracting the second <equation> from the first gives $10s_y=7$, so $s_y=7/10$ and $s_x=13/10$. Hence
$$
\boxed{(x_1,y_1)=\left(\frac3{10},-\frac3{10}\right).}
$$
The <gradient> there is
$$
\nabla F(3/10,-3/10)=\begin{pmatrix}0\\147/100\end{pmatrix}\ne\mathbf0.
$$
\b[A <critical point> has not been found after one iteration.]
At the alternative starting <point> $(0,2/3)$, the <Hessian matrix> and <gradient> are
$$
\nabla^2F=\begin{pmatrix}2&2\\2&2\end{pmatrix},\qquad
\nabla F=\begin{pmatrix}4/3\\0\end{pmatrix}.
$$
The correction would need to satisfy $2s_x+2s_y=-4/3$ and $2s_x+2s_y=0$ simultaneously. This <linear system> is inconsistent. \b[The standard iteration cannot take even its first step]: the <Hessian matrix> is a <singular matrix>, and the required correction does not exist. This is a <Newton correction compatibility condition> failure, not merely an ambiguity between several corrections.
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