If $a$ is a regular value of a function $g:\mathbb{R}^3\rightarrow \mathbb{R}$, we define the regular surface $S = g^{-1}(a)$. If U is an open set of $\mathbb{R}^2$ and $x:U\rightarrow \mathbb{R}^3$ is a differentiable, homeomorphic map such that for each $q\in U$, the differential $dx_q:\mathbb{R}^2\rightarrow \mathbb{R^3}$ is one-to-one (that is, $x$ satisfies the conditions for a parametrization of $S$), is $x$ automatically a parametrization of $S$?
I think the answer is no, but how the question is asked is a little bit ambiguous for me. We need that for each $p\in S$, then the range of the map $x$ needs to be $V\cap S$ where $V$ is a neighborhood of $p$. Otherwise, we could have that $x$ does not intersect the surface $S$. Am I overcomplicating things?