A surface is locally the graph of a function

We know that any surface is locally the graph of a function. In fact, we proved two things. First, the graph is defined in one of the three coordinate planes and, second, we know exactly what plane is. The proof says that if $p\in S$ and $X$ is a parametrization around $p$, then the rank of the derivative $dX_q$, $q=X^{-1}(p)$, is $2$. In particular, a $2\times 2$ matrix of $$dX_q=\left(\begin{array}{ccc}x_u& y_u&z_u\\ x_v& y_v&z_v\end{array}\right)$$ has non zero determinant, where $X(u,v)=(x(u,v),y(u,v),z(u,v))$. If we choose $$\left|\begin{array}{cc}x_u& y_u\\ x_v&y_v\end{array}\right|(q)\not=0,$$ then it is a graph on the $xy$-plane. 

It is clear that if $S$ is a graph on the $xy$-plane around $p$, then $T_pS$ can not be vertical, that is, $T_pS$ can not be orthogonal to the $xy$-plane. This can also proved in terms of the projection map. Let $\pi:{\mathbb R}^3\rightarrow{\mathbb R}^2$ be the projection $\pi(x,y,z)=(x,y)$ onto the $xy$-plane. This map is smooth, so when we restrict to $S$ is a differentiable map, which we denote by $\pi$ again. If $S$ is a graph on the $xy$-plane, then $\pi$ would be a local diffeomorphism. Indeed, the differential of $\pi$ is $$(d\pi)_p(v)=(\pi\circ\alpha)'(0)=(\alpha_1'(0),\alpha_2'(0))=(v_1,v_2)$$ where $\alpha$ is a curve that represents $v\in T_pS$ and $\alpha(t)=(\alpha_1(t),\alpha_2(t),\alpha_3(t))$. We ask when $(d\pi)_p$ is an isomorphism. Thus if $v\in T_pS$ where $(d\pi)_p(v)=(0,0)$ then $v_1=v_2=0$. This means that $v=(0,0,v_3)$ is a vertical vector. We conclude that $(d\pi)_p$ is an isomorphism if and only if  the tangent plane is not orthogonal to the $xy$-plane. In such a case, the inverse function theorem asserts the existence of an open set $V\subset S$ around $p\in S$ and and open set $W\subset{\mathbb R}^2$ around $\pi(p)$ such that $\pi:V\rightarrow W$ is a diffeomorphism. If $\phi=\pi^{-1}:W\rightarrow V$ and we stand for $\phi=(\phi_1,\phi_2,\phi_3)$, then for any $(x,y,z)\in V$ we have $$(x,y,z)=\phi(\pi(x,y,z))=\phi(x,y)=(\phi_1(x,y),\phi_2(x,y),\phi_3(x,y)).$$ This proves that $$V=\mbox{graph}(f)=\{(x,y,f(x,y)):(x,y)\in W\},$$ where $f=\phi_3:W\rightarrow\mathbb{R}$.