= Solution
{c}
Assume $f,F$ are <continuously differentiable> near the point and $F_z\ne0$. The <implicit function theorem> supplies a local <function> $z(x,y)$ with
$$
z_x=-F_x/F_z,\qquad z_y=-F_y/F_z.
$$
At an interior <extremum> of $g(x,y)=f(x,y,z(x,y))$, the <multivariable chain rule> gives
$$
0=g_x=f_x-f_zF_x/F_z,\qquad0=g_y=f_y-f_zF_y/F_z.
$$
Multiply these identities by $F_y$ and $F_x$, respectively, and subtract. The terms containing $f_z$ cancel, giving
$$
\boxed{f_xF_y-f_yF_x=0.}
$$
No assumption $f_z\ne0$ is needed for this elimination; dividing by $f_z$ would unnecessarily exclude valid cases. If $F_z=0$, this particular local graph argument is unavailable and a different chart or the full <Lagrange multiplier> condition is needed.
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