Consider implicit function theorem. Since the conditions the theorem gives are only sufficient and not necessary I would like to know how can one prove that it is impossible to express (locally) one variable as a function of other variables in a equation.
Here is what I found in 2D: In 2D consider $f(x,y)=c$ (with $f\in C^1$) and a point $(x_0,y_0)$ such that $f(x_0,y_0)=c$. If $\frac{\partial f}{\partial y}|_{\{x_0,y_0\}}=0$ this is not enough to prove that it is impossible to express $y=\Psi(x)$ locally.
But if I find out that $\frac{\mathrm{d}}{\mathrm{d}x} f(x,\Psi(x))|_{x_0}\neq0$ then I can conclude that is impossible to express $y=\Psi(x)$ locally.
(EDIT: Here I'm assuming that in the calculations of $\frac{\mathrm{d}}{\mathrm{d}x} f(x,\Psi(x))|_{x_0}$ I only need to know that, if it was possible to express $z$, then it would be $\Psi (x_0)=y_0$. In the expression of $\frac{\mathrm{d}}{\mathrm{d}x} f(x,\Psi(x))|_{x_0}$ both $\Psi (x_0)$ and $\Psi' (x_0)$ would appear (and I don't know $\Psi' (x_0)$) but I'm assuming $\Psi' (x_0)$ vanishes and only $\Psi (x_0)$ remains.)
But how can I extend this in 3D?
Consider $f(x,y,z)=c$ (with $f\in C^1$) and a point $(x_0,y_0,z_0)$ such that $f(x_0,y_0,z_0)=c$. If $\frac{\partial f}{\partial z}|_{\{x_0,y_0,z_0\}}=0$ this is not enough to prove that it is impossible to express $z=\Phi(x,y)$ locally.
But if I find out that $\nabla f(x,y,\Phi(x,y))|_{\{x_0,y_0\}}\neq(0,0)$ then can I conclude that is impossible to express $z=\Phi(x,y)$ locally?
(EDIT: Again I'm assuming that in the calculations of $\nabla f(x,y,\Phi(x,y))|_{\{x_0,y_0\}}$ I only need to know that, if it was possible to express $y$, then it would be $\Phi (x_0,y_0)=z_0$. In the expression of $\nabla f(x,y,\Phi(x,y))|_{\{x_0,y_0\}}$ there would be $\Phi (x_0,y_0)=z_0$ but also the partial derivativers of $\Phi$, (and I don't know them) but I'm assuming the partial derivatives vanish and only $\Phi (x_0,y_0)$ remains.)