 Solution

Consider the principal congruence subgroupIt has finite index because $S{L}_2({\mathbb{F}}_3)$ is a finite group. It is torsion-free: by part (b), a finite-order element is conjugate to $A^i$ or $B^j$, while the reductions of $A$ and $B$ modulo $3$ still have orders $6$ and $4$, respectively. Hence no nonidentity power in either vertex group reduces to the identity.

It follows that $\Gamma (3)$ intersects every conjugate of the two vertex stabilizers trivially, so its action on the Bass-Serre tree is free. A group with a free group action on a tree is a free group. Consequently $S{L}_2(\mathbb{Z})$ contains the free subgroup $\Gamma (3)$ of finite index; equivalently, it is a virtually free group.