$M$ is a smooth manifold if and only if $f$ is a smooth function. Let $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$
Let $M:=\{(x,f(x))\in \mathbb{R}^{n+m}:x\in \mathbb{R}^m\}$
Claim is that $M$ is a smooth manifold if and only if $f$ is a smooth function.
My attempt:
Proof$:(\Leftarrow)$
Assume $f$ is a smooth function. $M$ is a graph.
Fix a point $c\in M$. By definition $c=(x,f(x))$ for some $x\in \mathbb{R}^m$
Set $U:=\mathbb{R}^m$
Set $V:=\mathbb{R}^{m+n}$
Set $h:V \rightarrow U$. $h$ is smooth because it is linear.
$$(x,y)\in \mathbb{R}^m \times \mathbb{R}^n \mapsto h(x,y)=x$$
The inverse is $g: U \rightarrow V$
$$x \mapsto (x,f(x))$$
Is smooth by assumption.
$h \circ g= id$ (Since it is a function composed with its inverse)
$g \circ h= id$ (Since it is a function composed with its inverse)
Therefore $h$ is bijective.Thus $M$ is a smooth manifold.
$(\Leftarrow)$
Assume $M$ is a smooth manifold... Not sure where to go from here...
 A: This is false as written. $M$ will be a smooth manifold if $f$ is merely continuous, so $f$ need not be smooth.
To see this, note that we have two maps $\Phi$ from $\mathbb{R}^n$ to $M$ defined by $x \mapsto (x,f(x))$ and a map $\Psi$ from $M$ to $\mathbb{R}^n$ defined by $(x,f(x)) \mapsto x$. These maps are continuous and are mutually inverse, and so it is clear that $M$ is homeomorphic to $\mathbb{R}^n$ itself. If you require that $M$ have neighborhoods which are balls rather than all of Euclidean space, just compose with such a homeomorphism.
We have cheated the problem away by finding an atlas of one chart, so that points where $f$ may not be smooth are not relevant. This works on many familiar surfaces, such as the cone. 
However, it should be noted that $M$ will not be a smooth $\textit{submanifold}$ of $\mathbb{R}^{m+n}$.
To see that $M$ is not a smooth submanifold of $\mathbb{R}^{m+n}$, recall that $M$ is a smooth submanifold if it is a manifold in its own right, and there is an embedding into the ambient manifold (i.e. an injective smooth map with injective derivative).
Now given any atlas on $M$ (maybe take the easy cheater's atlas of one chart), we can try to make an embedding of $M$ into $\mathbb{R}^{n+m}$. Here we can see what goes wrong - if $f$ is not smooth at a point, any candidate for an embedding cannot be smooth at whatever point $f$ is not smooth at. This means that there cannot be such an embedding.
