Rigorous proof of surface area formula The surface integral over surface $S$ (which is given by $z=f(x,y)$, where $(x,y)$ is a point from the region $D$ in the $xy$-plane) is:
$$ \iint\limits_{S} g(x,y,z)\ dS = \iint\limits_{D} g(x,y,f(x,y))\ {{\sqrt {\,{{\left[ {{f_x}} \right]}^2} + {{\left[ {{f_y}} \right]}^2} + 1} \,dA}}$$
Does there exist a rigorous proof of this formula in real analysis.

There are various intuitive arguments showing that your formula, or the standard formula for the computation of surface area, is correct, and gives the expected results. But a full proof necessitates exact definitions of "surface" and "area", and the invariance of area under orthogonal mappings, etcetera.
If ${\rm d}S$ is a tiny piece of your surface $S$, located at the point $p\in S$, and this piece is projected orthogonally to the $(x,y)$-plane, then we obtain there a tiny "area element" ${\rm d}A$. The area scaling factor between these two pieces is $\cos\phi$, whereby $\phi$ is the angle between the tangent plane of $S$ at $p$ and the $(x,y)$-plane: 
$${\rm d}S={1\over\cos\phi}\,{\rm d}A\ .$$
It is easy to verify that $${1\over\cos\phi}=\sqrt{1+f_x^2+f_y^2}$$
when $S$ is given in the form $z=f(x,y)$.
