Assume that the equation $F(x,y,p)=0$ defines a regular submanifold $M$ of $R^3$. Consider the projection $\pi :M \rightarrow R^2$, given by $\pi (x,y,p)=(x,y)$.
By the implicit function theorem, in a neighborhood of a regular point of this map, $M$ is the graph of a smooth function $p=v(x,y)$. This is the part that I don't understand, since I would apply the inverse function theorem in the following way:
let $U$ be an open subset of $R^2 \times R$, consider the smooth function $F :U \rightarrow R$, if there exists a point $(x_0,y_0,p_0)\in U$ where $F(x_0,y_0,p_0)=0$ and $\partial F / \partial p (x_0,y_0,p_0)\neq 0$, then there exists a neighborhood $A \times B$ of $(x_0,y_0,p_0)$ in $U$ and a unique function $v:A \rightarrow B$ such that in $A \times B$,
$F(x,y,p)=0$ if and only if $p=v(x,y)$.
So my problem would be solved if I knew why $\partial F / \partial p (x_0,y_0,p_0)\neq 0$, where $(x_0,y_0,p_0)$ is a regular point of $\pi$. Thanks