Let $F: A \to B$ and $ G: B \to A $. Then
$ G = F^{-1}$ iff $G \circ F = I_A$ and $F \circ G = I_B$.
My Solution:
Now to prove if two functions are equal we have to show that both their domains are equal and all the elements in their domains have the same values in their co-domains.
(a) Let us take $G = F^{-1}$. Now domain of $(G \circ F = I_A) =$ Domain of $F$ = A = Domain of $ I_A$. Same could be done with the other statement. Now let us take an element $x$ to be an element of A. Then there exists a $y$ such that $f(x) = y$. So $g[f(x)]=g(y)$ is equal to $x$ because $(x,y)$ belongs to $f$ then $(y,x)$ belongs to $f^{-1}$. So $g[f(x)]=g(y) = x = I_A$ Same could be done with the other statement.
(b) Now let us take $G \circ F = I_A$ and $F \circ G = I_B$. As $I_A$ is one-to-one $F$ is one-to-one. And as $I_B$ is onto $B$, $F$ is onto $B$. Now this is where I get stuck. I don't know how will I be able to show $ G = F^{-1}$.
In this answer I put aside what your definition of inverse function is (see the comments) and I presume in (a) it was shown that $F^{-1}\circ F=1_{A}$ and $F\circ F^{-1}=1_{B}$.
Now suppose that $G\circ F=1_{A}$ for some function $G:B\rightarrow A$.
Then $G=G\circ1_{B}=G\circ F\circ F^{-1}=1_{A}\circ F^{-1}=F^{-1}$.