Given the following transformation $R(z)=\frac{3z-2}{2z-1}.$ With the convention that $R^n(z)=R(R(\cdots R(R(z))))$ $n$-times. I understand by induction, we get that $R^n(z)=\frac{(2n+1)z-2n}{2nz-(2n-1)}=1+\frac{z-1}{2nz-(2n-1)}.$ As $n \to \infty$ we get $R^n(z) \to 1$. So this has a fixed point of $1.$
Consider the following. If $R$ has a fixed point $\xi$, let $g(z)=\frac{1}{z-\xi}$ and let $S(z)=g(z)R(z)g^{-1}(z)$. $S$ will fix $z$ provided that $z=\infty$. Hence $S^n(z)\to\infty$ as $n\to\infty$ and $S^n(z)=g(z)R^n(z)g^{-1}(z).$ $g$ and $S^n(z)$ are explicitly known.
Where I'm stuck is the following. Consider $R^n(z)=g^{-1}(z)S^n(z)g(z)$. I'm not sure how to verify the form of $R^n(z)$ by using the facts for $S(z)$ and $g(z)$. I'm fairly confident that $g(z)=\frac{1}{z-1}$. I'm not sure how to find an $S^n(z)$ such that we will be able to work out $R^n(z)=\frac{(2n+1)z-2n}{2nz-(2n-1)}=1+\frac{z-1}{2nz-(2n-1)}.$
Note that $R^n(z)=g\circ S^n\circ g^{-1}(z)$ (not $g(z)S^n(z)g^{-1}(z)$). As the fixed point of $R$ is $1$, then $g(z)$ must be $1/(z-1)$. We find that if $S(z)=z+1$ then $$R^n(z)=g^{-1}(S^n(g(z)))=g^{-1}\left(S^n\left(\frac1{z-1}\right)\right) =g^{-1}\left(\frac1{z-1}+n\right) =g^{-1}\left(\frac{nz-n+1}{z-1}\right).$$ But $g^{-1}(z)=(1+z)/z$, so $$R^n(z)=\frac{(n+1)z-n}{nz-n+1}.$$ As this disagrees with $R^n$ with $n=1$, then $S$ could not have been $S(z)=z+1$. I think actually $S(z)=z+2$. Repeating this calculation with the correct $S(z)$ should give your formula for $R^n(z)$.