It's well known that if a function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ ( possible multivariable) is smooth, then it's graph $((x,y) \in \mathbb{R}^{n+m}: y = f(x))$ is a smooth manifold. This follows from a simple usage of inverse function theorem.
My question is if the reverse is true, i.e, if the graph is a smooth manifold, does it imply that the map is smooth in itself?
No. Consider the graph of $y=x^{1/3}$, the manifold is smooth but the function is not. Admittedly there does exist a smooth function for this manifold, but I'm assuming that you've already partitioned the dimensions into $y$ and $x$ prior to constructing the function.