 Vector area element of an implicit surface

For a regular implicit surface $f(x,y,z)=0$ with $f_z\ne 0$, the implicit function theorem gives a local graph $z=h(x,y)$. Its upward vector area element is $(-h_x,-h_y,1)dxdy$. Differentiating $f(x,y,h(x,y))=0$ yields $h_x=-f_x/f_z$, $h_y=-f_y/f_z$, proving the formula. Division by $f_z$ makes the final component positive independently of the sign of the original gradient. At a point where $f_z=0$, another projection or parametrization is needed.