= Solution
Where $f_z\ne0$, the <implicit function theorem> makes the surface a local graph $z=h(x,y)$ with $h_x=-f_x/f_z$ and $h_y=-f_y/f_z$. Its upward <vector area element> is $(-h_x,-h_y,1)dx\,dy$. Therefore the <vector area element of an implicit surface> is
$$
\boxed{d\mathbf S=\frac{\nabla f}{f_z}\,dx\,dy,\qquad\mathbf F=\frac{\nabla f}{f_z},\quad\mathbf F\cdot\mathbf k=1.}
$$
Dividing by $f_z$ fixes the upward orientation even if the initially chosen gradient points downward. The restriction excludes a locally vertical graph, where this particular projected-area representation cannot be used.
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