If $(f \circ g)(x) = x$, and $(g \circ f)(x) = x$, then $g$ is the inverse of $f$. If $f: A\to B$ is bijective, then it has an inverse $g: B \to A$ defined as
\begin{equation} 
g = \left \{ \big( f(a), a \big) : a \in A \right \}
\end{equation}
If $g: B \to A$ is any function for which,
\begin{equation}
\begin{aligned}
(f \circ g)(x) & = x, & \forall x \in B, \\  
(g \circ f)(x) & = x, & \forall x \in A.  
\end{aligned}
\end{equation}
then $g \equiv f^{-1}$ (inverse)
I basically need to show that if any two functions are composed with $x$ ( I don't if that's the proper way to say it) and you get back $x$, then those two functions are inverses of one another.
In other words  $g \equiv f^{-1}$ (inverse)
Any help with this would be really appreciated. I've been trying to get this down for a couple of days now.

Well, if you have a set-theoretic definition of the inverse, it makes sense to expand the other two statements that you have in terms of set theory as well. In particular, we can expand the statement
$$(f\circ g)(x)=x$$
to say that for any pair $(x,y)\in f$ we have that $(y,x)\in g$. This comes essentially directly from the definition of a composition of relations. The statement
$$(g\circ f)(x)=x$$
is essentially the converse to this - saying that $(y,x)\in g$ implies $(x,y)\in f$.
Putting these together gives the following: $(y,x)\in g$ if and only if $(x,y)\in f$. Given that the latter condition is equivalent to saying $y=f(x)$ we get that $(y,x)\in g$ if and only if $y=f(x)$ - which immediately gives that $g=\{(f(x),x):x\in A\}$.
