Let $\mathbb{R}_+:=(0,\infty)$. Let $f\colon \mathbb{R}_+\times\mathbb{R}\to\mathbb{R}$ be a $C^1$-function with $0$ as a regular value (i.e. $Df(0)\colon\mathbb{R}_+\times\mathbb{R}\to\mathbb{R}$ is surjective), and let $U=f^{-1}(0)\subset\mathbb{R}^2$. $U$ is a differentiable submanifold of $\mathbb{R}^2$.
Why is $M:=\{(x,y,z)\in\mathbb{R}^3 \mid f(\sqrt{x^2+y^2},z)=0\}$ a differentiable 2-dim. submanifold of $\mathbb{R}^3$?
To check this, I want to write $M$ as a zero zet of a $C^1$-function $g \colon\mathbb{R}^3\to\mathbb{R}$ with $\operatorname{rank}(\nabla g(a))=1$ for all $a\in M$, so that it follows that it is a diff. submanifold with dimension $2$.
The canonical $g$ is $g \colon\mathbb{R}^3\to\mathbb{R}$, defined by $g(x,y,z)=f(\sqrt{x^2+y^2},z)$. If I am not mistaken, it is $\nabla g(x,y,z)=(2x\frac{\frac{\partial f(\sqrt{x^2+y^2},z)}{\partial x}}{\sqrt{x^2+y^2}}, 2y\frac{\frac{\partial f(\sqrt{x^2+y^2},z)}{\partial y}}{\sqrt{x^2+y^2}}, \frac{\partial f(\sqrt{x^2+y^2},z)}{\partial z})$. Now, $\operatorname{rank}(\nabla g(a))=1$ for all $a\in M$ if $\nabla g(a)\neq 0$ for all $a\in M$.
Is it correct until now? How to proceed?
$g$ is the composition of two function $h(x,y,z)=(\sqrt{x^2+y^2},z)$ and $f$,
$dh(x,y,z)=\pmatrix{{{2x}\over{\sqrt{x^2+y^2}}}&{{2y}\over{\sqrt{x^2+y^2}}}&0\cr 0&0&1}$ and $\pmatrix{{{\partial f}\over {\partial x}}&{{\partial f}\over {\partial y}}}$
It composition is $\pmatrix{{{2x}\over{\sqrt{x^2+y^2}}}{{\partial f}\over {\partial x}}&{{2y}\over{\sqrt{x^2+y^2}}}{{\partial f}\over {\partial x}}& {{\partial f}\over {\partial y}}}$ which is a submersion on $M-\{(0,0,0)\}$.