If I want to use the Implicit function theorem for a function $f:\mathbb{R}^3\rightarrow \mathbb{R}$. Is the order of the variables important? What I mean with that is, if I can solve $f(x,y,z)$ for $y$?
Thoughts:
Assuming $f(a,b,c)=0$ for $(a,b,c)\in\mathbb{R}^3$ and $\frac{\partial f}{\partial y}(a,b)\neq 0 $ then* $f$ is locally solvable for $y$ i.e. there exist a neighborhood $U\subset \mathbb{R}^2$ of $(a,c)$ and a continuous differentiable function $g: U\rightarrow \mathbb{R}$ with $y=g(x,z)$ and $f(x,g(x,z),z)=0$ for every $(x,z)\in U$
But can I use the theorem * like that?
The general form in the theorem is given by: $f:\mathbb{R}^m\times \mathbb{R}^n\rightarrow \mathbb{R}^m$
$f:\mathbb{R}^3\rightarrow \mathbb{R}$ can be written as:
a) $f:\mathbb{R}^m\times \mathbb{R}^n\rightarrow \mathbb{R}^m$ with $m=1,n=2$ if I want to search for a solution $x=g(y,z)$
b)$f:\mathbb{R}^n\times \mathbb{R}^m\rightarrow \mathbb{R}^m$ with $m=1,n=2$ if I want to search for a solution $z=g(x,y)$.
c) How can I write $f:\mathbb{R}\times\mathbb{R}^m\times\mathbb{R}\rightarrow \mathbb{R}^m$ if I want to search for a solution $y=g(x,z)$ with $m=1$? I cannot identify $n$.
I'm having some trouble phrasing my question, hopefully it still makes sense to you.
The order of the variables doesn't matter. If it helps, you can just define a new function with the variables in the order you want and then apply the implicit function theorem to the new function: $$ F(x,z,y) := f(x,y,z).$$