Let $F:\mathbb{R}^2\to\mathbb{R}$ of class $C^2$ and $\displaystyle\frac{\partial f}{\partial v}(u,v)\neq0\; \forall(u,v)\in\mathbb{R}^2$. If $(x_0,y_0,z_0)\in\mathbb{R}^3$ is such that $f(x_0y_0,z_0-2x_0)=0$ show that can be defined a function $z=f(x,y)\iff F(xy,z-2x)=0$ for some open neighborhood of $(x_0,y_0).$
I think this may be solved this way: $\displaystyle\frac{\partial f}{\partial (z_0-2x_0)}(x_0y_0,z_0-2x_0)\neq0$, let $U$ and $V$ be open neighborhoods of $x_0y_0$ and $z_0-2x_0$ then by the implicit function theorem can be defined a function $g:U\to V$ such that $g(xy)=z-2x$ which means that $z=g(xy)+2x$. Defining a new function $f$ by $f(x,y)=g(xy)+2x$ seems to solve the problem.
Is this correct?. I'm not sure if I get the full implications of $\displaystyle\frac{\partial f}{\partial v}(u,v)\neq0$.
Consider the function $G$ defined on $\mathbb{R}^{2}$ by :
$$ \forall (x,y) \in \mathbb{R}^{3}, \; G(x,y,z) = F(xy,z-2x) $$
Then, there exists $(x_{0},y_{0},z_{0}) \in \mathbb{R}^{3}$ such that : $G(x_{0},y_{0},z_{0})=0$ and :
$$ \frac{\partial G}{\partial z}(x,y,z) = \frac{\partial F}{\partial v}(xy,z-2x) $$
So that $\displaystyle \frac{\partial G}{\partial z}(x_{0},y_{0},z_{0}) = \frac{\partial F}{\partial v}(x_{0}y_{0},z_{0}-2x_{0}) \neq 0$. So, applying the implicit function theorem to $G$ gives : there exists an open neighborhood $V$ of $(x_{0},y_{0})$ in $\mathbb{R}^{2}$, an open neighborhood $W$ of $z_{0}$ in $\mathbb{R}$ and a function $f \, : \, V \, \longrightarrow \, W$, $\mathcal{C}^{1}$ on $V$, such that :
$$ \Big( \, \forall (x,y,z) \in V \times W, \; G(x,y,z)=0 \, \Big) \Leftrightarrow \Big( \, \forall (x,y) \in V, \; z=f(x,y) \, \Big) $$