Consider $M =\mathbb{Z}/3\mathbb{Z}$ as a $R = \mathbb{Z}/9\mathbb{Z}$ module. I'm trying to calculate what $Ext_{R}^n(M,M)$ is for all $M$. To this end, I let
$$\cdots \xrightarrow{\times 3} \mathbb{Z}/9\mathbb{Z} \xrightarrow{\times 3} \mathbb{Z}/9\mathbb{Z} \xrightarrow{\times 3}\mathbb{Z}/9\mathbb{Z} \xrightarrow{\text{projection}} \mathbb{Z}/3\mathbb{Z} \rightarrow 0$$
be a free (and hence projective) resolution. To calculate $Ext_{R}^n(M,M)$, I now just take the homology groups of $$0 \rightarrow \text{Hom}(\mathbb{Z}/3\mathbb{Z},\mathbb{Z}/3\mathbb{Z})\xrightarrow{\times 3}\text{Hom}(\mathbb{Z}/9\mathbb{Z},\mathbb{Z}/3\mathbb{Z})\xrightarrow{\times 3}\text{Hom}(\mathbb{Z}/9\mathbb{Z},\mathbb{Z}/3\mathbb{Z})\xrightarrow{\times 3} \cdots.$$ Since $\text{Hom}(R,M) \cong M$, the above is just the chain $$0\rightarrow \mathbb{Z}/3\mathbb{Z} \xrightarrow{\times 3} \mathbb{Z}/3\mathbb{Z} \xrightarrow{\times 3} \mathbb{Z}/3\mathbb{Z}\xrightarrow{\times 3} \cdots$$ so that the kernels are all of $\mathbb{Z}/3\mathbb{Z}$ and the images are just $0$ so that $Ext_R^N(M,M) = \mathbb{Z}/3\mathbb{Z}$ for all $n$. Is this correct, have I made a minor error, or have I fundamentally misunderstood something important? (Or both!)
You made a minor mistake: note that if you have a right exact functor $F$, and you want to compute $L_*F(X)$, you take a projective resolution $P_*\to X$ and then take the homology of $F(P_*)$, not the homology of $F(P_*\to X)$.
So your minor mistake is saying that you're looking for the homology groups of $\hom(\mathbb Z/3,\mathbb Z/3)\to \hom(\mathbb Z/9,\mathbb Z/3) \to \hom(\mathbb Z/9,\mathbb Z/3) \to \dots$
(note, by the way, that the first map would be identified with the identity $\mathbb Z/3\to\mathbb Z/3$, not with $\times 3$, as it is the dual of the projection $\mathbb Z/9\to\mathbb Z/3$)
The chain complex whose homology you're looking for only has $\hom(\mathbb Z/9,\mathbb Z/3)$'s in it. It does, however, give you the result you say.