The graph of a smooth real function is a submanifold Given a function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m $ which is smooth, show that
$$\operatorname{graph}(f) = \{(x,f(x)) \in \mathbb{R}^{n+m} : x \in \mathbb{R}^n\}$$
is a smooth submanifold of $\mathbb{R}^{n+m}$.
I'm honestly completely unsure of where or how to begin this problem. I am interested in definitions and perhaps hints that can lead me in the right direction.

This is an application of the Implicit Function Theorem. You have a function $f : \mathbb{R}^n \to \mathbb{R}^m$ and you construct the graph given by: $\{ (x,y) \in \mathbb{R}^n \times \mathbb{R}^m : y = f(x) \}.$ Let me define a new map, say, $G : \mathbb{R}^{n+m} \to \mathbb{R}^m$ given by $G : (x,y) \mapsto y-f(x).$ I have defined this the way I have so that the graph of $f$ is the zero-level set of $G$, i.e. the graph of $f$ is the set of $(x,y) \in \mathbb{R}^n \times \mathbb{R}^m$ such that $G(x,y) = 0.$ 
In brutal detail this map is really:
$$G : (x_1,\ldots,x_n,y_1,\ldots,y_m) \mapsto (y_1-f_1(x_1,\ldots,x_n),\ldots,y_m-f_m(x_1,\ldots,x_n)) \, .$$
We need to calculate the Jacobian Matrix of $G$. A quick calculation will show you that:
$$ J_G = \left[\begin{array}{c|c} -J_f & I_m \end{array}\right] ,$$
where $J_f$ is the $m \times n$ Jacobian matrix of $f$ and $I_m$ is the $m \times m$ identity matrix. The matrix $J_G$ is an $m \times (m+n)$ matrix.
To be able to apply the IFT, we need to show that $0$ is a regular value of $G$. (After all, the graph of $f$ is $G^{-1}(0).$) We can do this by showing that none of the critical points get sent to 0 by $G$. Notice that $G$ has no critical points because $J_G$ always has maximal rank, i.e. $m$. This is clearly true since the identity matrix $I_m$ has rank $m$.
It follows that the graph of $f$ is a smooth, parametrisable $(n+m)-m=n$ dimensional manifold in a neighbourhood of each of its points.
