So I have this practice problem. I see that $x^3 - x - 1$ is irreducible over $Z/3$ and so I guess the number of elements in $R$ would just be the same as a degree $3$ field extension over $F_3$ so $27$ elements I think? Then since its irreducible it must also be a maximal ideal in $ \space F[x]$ which is a PID and so the quotient ring is a field which must be isomorphic to $F_{27}$ I think? So I guess the splitting field would be $x^{3^3-1}-x$, but I don't know how to see which roots of unity which has. I'm not even sure if what I have so far is correct
You're on the right track.
Since $R=\mathbb F_{27}$ is a field, the multiplicative group $R^\times$ is cyclic of order $26$. Therefore, it has elements of all orders dividing $26$. These are the primitive roots of unity in $R$.