Submanifold concerning the map of a function Let $f: \mathbb{R}^+ \times \mathbb{R} \rightarrow \mathbb{R}$ with $f \in C^1$ and the Jacobian of $f$ has full rank (1) for all $z \in \mathbb{R}^+ \times \mathbb{R}$. 
Then $M=\{z=(z_1,z_2,z_3) \in \mathbb{R}^3|(z_1,z_2)\neq (0,0),f(\sqrt{z_1^2+z_2^2},z_3)=0\}$ is a 2-dimensional submanifold of $\mathbb{R}^3$.
I am stuck on this task, I have some approach but I couldn't quite finish my proof (and I am especially not sure if I did any mistake):
My idea was to write $M$ as the roots of a function $g=f*h$ with a Jacobian of maximal rank (it follows that M is a submanifold).
Let use take $h: \mathbb{R}^2\backslash\{0\} \times \mathbb{R} \rightarrow \mathbb{R}^+ \times \mathbb{R}$ with $h(z_1,z_2,z_3)=(\sqrt{z_1^2+z_2^2},z_3)$ we get that $h$ is continuous because both components are continuous and therefore the composition $g=f*h$ is continuous.
And therefore it follows that $$J(f*h)=\left(
\begin{array}{ccc}
 \frac{z_1
   \partial_1f\left(\sqrt{z_1^2+z_2^2},\text{z3}\right)}
   {\sqrt{z_1^2+z_2^2}} & \frac{z_2
   \partial_1f\left(\sqrt{z_1^2+z_2^2},\text{z3}\right)}
   {\sqrt{z_1^2+z_2^2}} &
   \partial_2f\left(\sqrt{z_1^2+z_2^2},\text{z3}\right)
\end{array}
\right)$$
Now I think I am done if I can show that at least one entry of that matrix is always unequal to zero. Thank you for your help.

You can do this using regular values. I think the necessary machinary comes from the Regular Value Theorem but the idea is to show that 0 is a regular value of the image, that is f^{-1}(0) contains no critical points, but seeing as we're told that the jacobian has full rank (of 1) this must be true and hence M is a regular submanifold of codim 1. 
