= Solution
Parametrize the graph by $\mathbf r(x,y)=(x,y,f(x,y))$ over its projection $D$ onto the $xy$ plane. The tangent vectors are $(1,0,f_x)$ and $(0,1,f_y)$, so their <cross product> is $(-f_x,-f_y,1)$. Hence the scalar <surface element> is
$$
dS=|\mathbf r_x\times\mathbf r_y|\,dx\,dy=\sqrt{1+f_x^2+f_y^2}\,dx\,dy,
$$
and
$$
\boxed{\operatorname{area}(S)=\iint_D\sqrt{1+f_x^2+f_y^2}\,dx\,dy.}
$$
The printed subscript $S$ is shorthand for integrating over the projected parameter region: $dx\,dy$ belongs to $D$, while the square-root factor converts projected area into actual surface area. The corresponding upward <vector surface element of a graph> is $(-f_x,-f_y,1)\,dx\,dy$.
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