Answer
We prove the formula:
${\rm{Area}}\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \dfrac{1}{{\left| {\cos \phi } \right|}}{\rm{d}}A$
Work Step by Step
The graph $z = g\left( {x,y} \right)$ has the parametrization:
$G\left( {x,y} \right) = \left( {x,y,g\left( {x,y} \right)} \right)$
So,
${{\bf{T}}_x} = \left( {1,0,{g_x}} \right)$
${{\bf{T}}_y} = \left( {0,1,{g_y}} \right)$
Recall Theorem 1:
${\rm{Area}}\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$
where ${\bf{N}}\left( {x,y} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y}$, a normal vector to the surface $S$.
${\bf{N}}\left( {x,y} \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&{{g_x}}\\
0&1&{{g_y}}
\end{array}} \right| = - {g_x}{\bf{i}} - {g_y}{\bf{j}} + {\bf{k}}$
Let the $z$-axis be the vertical axis. Since $\phi = \phi \left( {x,y} \right)$ is the angle between the normal to $S$ and the vertical. So,
${\bf{N}}\left( {x,y} \right)\cdot{\bf{k}} = ||{\bf{N}}\left( {x,y} \right)||\cos \phi $
$\left| {{\bf{N}}\left( {x,y} \right)\cdot{\bf{k}}} \right| = ||{\bf{N}}\left( {x,y} \right)||\left| {\cos \phi } \right|$
$||{\bf{N}}\left( {x,y} \right)|| = \dfrac{{\left| {{\bf{N}}\left( {x,y} \right)\cdot{\bf{k}}} \right|}}{{\left| {\cos \phi } \right|}}$
Substituting $||{\bf{N}}\left( {x,y} \right)||$ into ${\rm{Area}}\left( S \right)$ gives
${\rm{Area}}\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \dfrac{{\left| {{\bf{N}}\left( {x,y} \right)\cdot{\bf{k}}} \right|}}{{\left| {\cos \phi } \right|}}{\rm{d}}x{\rm{d}}y$
But
${\bf{N}}\left( {x,y} \right)\cdot{\bf{k}} = \left( { - {g_x}{\bf{i}} - {g_y}{\bf{j}} + {\bf{k}}} \right)\cdot{\bf{k}} = 1$
and $dA = dxdy$.
Therefore
${\rm{Area}}\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \dfrac{1}{{\left| {\cos \phi } \right|}}{\rm{d}}A$