Find the graph of a function Let A be an open of $\mathbb R^m$,with $m=n+k$ and $f:A \to \mathbb R^k$ a function of $C^1$.Let $c \in \mathbb R^k$ a regular value of $f$. Then, for every $x \in f^{-1}(c)$ there is an open $U$ of $\mathbb R^{n}$ and an open $V$ of $\mathbb R^k$ such that $f^{-1}(c) \cap (U \times V)$ is the graph of a function $g:U \to V$ of class $C^1$.
My attempt, 
let $f:A\to \mathbb R^k$ where $c$ is the regular value of $f$ ,then $\forall x \in f^{-1}(c)$ we obtain that $Df(x)$ is an injector. Applying the implicit function theorem, there are open $U \subset \mathbb R^n $,$V \subset \mathbb R^k $ with  $x_{1} \in U$, $x_{2} \in V$ (where $x=(x_{1},x_{2})$), and $U \times V \subset A$ such that 
$\forall x \in U $,$\exists$! $y=g(x) \in V$ with $f((x,g(x)))=c$ where $g:U \to V$ is $C^1$.
note that  $\{(x,g(x)) \in \mathbb R^{n+k}\} \subset U \times V$ then
$$f^{-1}(c) \cap (U \times V) \subset \{(x,g(x)) \in A \cap (U \times V) \} = \{(x,g(x)) \in  (U \times V)\}$$ It is necessary to prove equality?
 A: As you tried to do, what we need to do is to prove that $f^{-1}(c) \cap \left( U \times V \right)$ is the graph of the function $g$ whose existence is guaranteed by the implicit function theorem.
For that, note that $f^{-1}(c) \cap \left( U \times V \right) = \left\{(x,y)\in U \times V: f(x,y) = c\right\}$.
The implicit function theorem tells us that, for every $x \in U$, there is exactly one $y \in V$ such that $f(x,y)=c$, which is $y = g(x)$.
Therefore, we can write $f^{-1}(c) \cap \left( U \times V \right) = \left\{(x,g(x)): x \in U\right\}$, which is by definition the graph of $g$.
(Edit:) Answering your last question: you do need to prove equality, since it is asked to prove that this set is the graph of the function $g$, not just a subset of this graph.
You could argue, a subset of the graph of $g$ is also a graph of a function, it's the graph of the restriction of $g$ to a certain subset of its domain. However, this subset may not be open, and it's asked that the function be defined on an open set.
