The Implicit Function Theorem tells us that if we have a smooth function $f: \mathbb{R}^{n+k} \to \mathbb{R}^{k}$ such that $f(a) = 0$ for some $a \in \mathbb{R}^{n+k}$, then we can find points close to $a$ that lie in the same level set $f^{-1}(0)$ which can be expressed as the graph of a function $\mathbb{R}^{n} \to \mathbb{R}^{k}$, assuming that a $k$-dimensional restriction of $Df(a)$ mapping $\mathbb{R}^{k} \to \mathbb{R}^{k}$ is invertible.
Is the converse of this statement true? That is, if we assume the hypotheses of the Implicit Function Theorem, is it true that the graph of a function $\mathbb{R}^{n} \to \mathbb{R}^{k}$, as a subset of $\mathbb{R}^{n+k}$, can be locally expressed as the level set of some function $\mathbb{R}^{n+k} \to \mathbb{R}^{k}$? How would one prove this?
Consider a simple situation: We have a $C^1$ function $f:(0,1)\to \mathbb R.$ The graph of $f$ is the set $G=\{(x,f(x)):x\in (0,1)\}.$ On $U=(0,1)\times \mathbb R,$ define $g(x,y)= y-f(x).$ Then $g\in C^1(U),$ and the level set $\{g=0\}$ is precisely $G.$