First, we note that $f \circ g$ will be a map $A \to C$. If we assume that $f \circ g$ is bijective, we automatically have the following two observations:
- Surjectivity: for each $c \in C$, there exists $a \in A$ such that $(f \circ g)(a) = c$.
- Injectivity: if $(f \circ g)(a) = (f \circ g)(a^\prime)$, then we must have $a = a^\prime$.
All that is needed from here are the two properties above. I'll write out the injectivity part and leave the surjective part to you.
To check that $g$ is injective, we proceed directly using the definition. Let's assume that $g(a) = g(a^\prime)$ for two $a,a^\prime \in A$. Applying $f$ to both sides of this equation, we find that
(f \circ g)(a) = f(g(a)) = f(g(a^\prime)) = (f \circ g)(a^\prime)
whence we must have $a = a^\prime$ (why? which property is used here?). Using 1. in a similar way, we can show that $f$ must be surjective as well.
Also, the hypothesis is that $f \circ g$ is bijective. Consequently, you do not need to prove that $f \circ g$ is bijective as this property is given!