= Solution
{c}
For a twice <continuously differentiable function> $F$, its <gradient> and <Hessian matrix> are
$$
\nabla F(x,y)=\begin{pmatrix}F_x\\F_y\end{pmatrix},\qquad
H_F(x,y)=\begin{pmatrix}F_{xx}&F_{xy}\\F_{yx}&F_{yy}\end{pmatrix}.
$$
A <critical point> solves $\nabla F=0$. The <Jacobian matrix> of this vector equation is $H_F$, so the <multivariable Newton iteration> is
$$
\boxed{H_F(\mathbf x_k)\mathbf s_k=-\nabla F(\mathbf x_k),\qquad\mathbf x_{k+1}=\mathbf x_k+\mathbf s_k.}
$$
Equivalently $\mathbf x_{k+1}=\mathbf x_k-H_F(\mathbf x_k)^{-1}\nabla F(\mathbf x_k)$ when the <Hessian matrix> is invertible. Solving the <linear system> is the direct computational form. A <critical point> found in this way need not be a <local minimum>; its type requires a separate check.
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