Free subgroups of $PSL(2,\mathbb{Z})$ of index 6 There are two "natural" subgroups of $PSL(2,\mathbb{Z})\cong C_2\ast C_3$ of index 6. One is the congruence subgroup $\Gamma_0(2)$ which is the kernel of the map $PSL(2,\mathbb{Z})\to PSL(2,\mathbb{Z}/2\mathbb{Z})$. The other subgroup $H$ is the kernel of the map $C_2\ast C_3\to C_2\times C_3$. Here are two similarities between these two subgroups:


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*Both $\Gamma_0(2)$ and $H$ are subgroups of $PSL(2,\mathbb{Z})$ of index 6.

*Both $\Gamma_0(2)$ and $H$ are free groups of rank 2.


However, $PSL(2,\mathbb{Z}/2\mathbb{Z})\cong S_3$ and $C_2\times C_3\cong C_6$ so these are different subgroups of $PSL(2,\mathbb{Z})$. Moreover, $\Gamma_0(2)$ is freely generated by the matrices $\begin{bmatrix}1&2\\0&1\end{bmatrix}$ and $\begin{bmatrix}1&0\\2&1\end{bmatrix}$ whereas $H$ is freely generated by the matrices $\begin{bmatrix}2&1\\1&1\end{bmatrix}$ and $\begin{bmatrix}1&1\\1&2\end{bmatrix}$. This last statement can be seen by noting that if $a=\begin{bmatrix}0&-1\\0&1\end{bmatrix}$ generates $C_2$ and $b=\begin{bmatrix}-1&-1\\1&0\end{bmatrix}$ generates $C_3$ then $H$ is freely generated by $abab^2$ and $ab^2ab$.
What is going on here? More precisely,


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*Are these two subgroups the largest free subgroups of $PSL(2,\mathbb{Z})$?

*Are there any other free subgroups of $PSL(2,\mathbb{Z})$ of index 6?

*Is there any reason to expect that $PSL(2,\mathbb{Z})$ contains two free subgroups of rank 2 and index 6 with different quotients?

 A: Let $G = \langle x,y \mid x^2,y^3 \rangle \cong {\rm PSL}(2,\mathbb{Z})$.
Question 1. Yes. Let $H < G$, and consider the permutation action of $G$ on the (left or right) cosets of $H$ in $G$. If $|G:H| < 6$, then it is not possible for the images of both $x$ and $y$ to act fixed point freely, and so $H$ contains a  conjugate of $x$ or $y$ and hence cannot be free.
Question 2. No, but the two subgroups that you have found are the only two normal subgroups of index $6$ in $G$. You can see this by observing that there is essentially only one surjective group homomorphism from $G$ to each of $C_6$ and $S_3$ (i.e. up to equivalence under an automorphism of $C_6$ or $S_3$), so there are only two possible kernels $H$.
By a computer calculation, I found that there is also one conjugacy class of non-normal subgroups $H$ with $|G:H| = 6$ and with $H$ free of rank $2$, and a representative of this class is $H=\langle yx, y^{-1}(xy)^3 \rangle$. The quotient of $G$ by the core of $H$ is isomorphic to $S_4$, and there are three conjugates of $H$ in $G$. 
Question 3. I can only say here that the reason is that we have a proof that this is the case!
Note that the Kurosh Subgroup Theorem says that any subgroup of $G$ is a free product of conjugates of $\langle x \rangle$, $\langle y \rangle$ and a free subgroup of $G$. So, for $H \lhd G$, if $H$ does not contain $x$ or $y$, then it must be free. I believe that the rank of free subgroups of free products can be calculated using Euler Characteristics, but I don't know the details.
