I was recently introduced to a particular definition of a regular surface which is:
A (regular) surface is a subset $S \subset \mathbb{R}^{3}$ with the property that around every point $p \in S$ we can find a map $\varphi: U \rightarrow S$ with $U \subset \mathbb{R}^{2}$ an open subset and $p \in \varphi(U)$, that meets the following conditions:
i) $\varphi$ is a $C^{1}$-map on $U$,
ii) $\varphi$ maps $U$ bijectively onto its image $\varphi(U) \subset S$,
iii) For every $a \in U$ it follows that the derivative $\varphi'(a) \in \text{Hom}(\mathbb{R}^{2}, \mathbb{R}^{3})$ is injective.
I have tried to prove the following proposition:
If $f: U \subset \mathbb{R}^{3} \rightarrow \mathbb{R}$ is a $C^{1}$ function and $a \in f(U)$ is a regular value of $f$, then $f^{-1}(a) = \{(x, y, z) \in U \mid f(x, y, z) = a\}$ is a regular surface in $\mathbb{R}^{3}$
I have found a proof which makes use of an alternate definition of a regular surface which uses homeomorphisms. I know there is a way to prove that $f^{-1}(a)$ is a regular surface by my definition, but can't seem to find complete the proof.
I have already shown that the total derivative, which is a 3x3 matrix, has determinant not equal to zero and thus is invertible. I found that I have to use the inverse function theorem to do this but I'm not sure how. Thanks in advance!