Localization of an integer quotient is a field 
Let $R:=\mathbb{Z}/24\mathbb{Z}$ be our ring, $f: \mathbb{Z}\to R$ be the canonical quotient map (i.e. $f$ sends an element to its equivalence class) and $q$ be the ideal generated by $f(3 +24\mathbb{Z})$. I want to show that $R_q:=(R\setminus q)^{-1}R$ is a field. 

My Ideas:
Since $f(3 +24\mathbb{Z})=f(3)$, we get $q=3\mathbb{Z}/24\mathbb{Z}$, hence $q$ is a prime ideal. 
Moreover it's sufficient to show that $R_q$ without the zero element is a group with respect to $*$ ($*$ is the multiplication sign), because of $R_q$ is a ring (in particular unitial and commutative).
But now I'm really stuck, because I don't see exactly how the elements of $R_q$ look like without doing long explicit computation of the equivalence classes of $R_q$. 
edit: I did some corrections, I hope it's now more clear. 
 A: The title statement "localization of an integer quotient is a field" is incorrect in general.
For example when $n$ is not squarefree (as is the case for $24=2^3\cdot 3$) then the localization at the prime which isn't squarefree contains nonzero nilpotent elements, so it is not a field. In the $n=24$ case, localizing at $2$ demonstrates this: if you localize at the complement of $(2)$ then $3$ becomes a unit, but then $3\cdot 8=24=0$ implies $8=0$.
However, with what that teaches us, I think you'll be able to solve your problem :) With the special choice of localizing at $(3)$ in your example, things are going to work out.
Here is a good approach. After you do the localization at the set $S=R\setminus(3)$, then $S^{-1}R$ is a local ring with unique maximal ideal generated by $3$ right? If you show that this ideal is zero, then the localization is a commutative ring with only two trivial ideals (and that is a field.) 
Work to show that the ideal generated by $3$ in the localization is zero. Of course, the fact that $24=0$ and $2$ is a unit plays an important role in this.
