How to find the integral closure of $\mathbb{Z}_{(3)}$ in the field $\mathbb{Q}(\sqrt{-5})$?

The integral closure of $\Bbb Z$ in $K={\Bbb Q}(\sqrt{-5})$ is the ring of integers ${\cal O}_K={\Bbb Z}[\sqrt{-5}]$, the latter equality because $-5\equiv 3\bmod 4$.

As you observed, ${\Bbb Z}_{(3)}$ is the localization of $\Bbb Z$ at the ideal $(3)$. Then, it follows from general properties (see Atiyah-Macdonald, Ch 5) that the integral closure $A$ of ${\Bbb Z}_{(3)}$ in $K$ is the localization at the ideal $(3)$ in ${\cal O}_K$.

Since $3$ splits in $K$, $A$ has two maximal ideals, namely ${\frak m}_{\pm}=(1\pm\sqrt{-5})A$.

Here is a different and (perhaps) somewhat more elementary approach than Andrea's.

Let $R$ be the integral closure of $\mathbb{Z}_{(3)}$ in $\mathbb{Q}(\sqrt{-5})$. Clearly $R$ is a subring of $\mathbb{Q}(\sqrt{-5})$, and thus we are looking for a necessary and sufficient condition on $a,b \in \mathbb{Q}$ such that $a+b\sqrt{-5} \in R$.

Let $P(a,b)$ be the minimal polynomial of $a+b\sqrt{-5}$ over $\mathbb{Q}$, i.e., the unique monic polynomial with $\mathbb{Q}$-coefficients of least degree satisfied by $a+b\sqrt{-5}$. Certainly if $P(a,b)$ has coefficients lying in $\mathbb{Z}_{(3)}$ then $a+b\sqrt{-5}$ lies in $R$. In fact the converse is true because the ring $\mathbb{Z}_{(3)}$ is integrally closed (e.g. it is a PID and PID $\implies$ UFD $\implies$ integrally closed). For a proof of this fact, see e.g. the section on Integrally Closed Domains in these notes. (Currently this is Theorem $260$ in $\S 14.5$, but that more precise information is subject to change.)

Try out this computation for yourself: you should get that $a+b\sqrt{-5} \in R \iff a,b \in \mathbb{Z}_{(3)}$.

Now in my notes I also prove the compatibility of integral closure with localization which Andrea uses in his answer: in fact it comes about five pages earlier than the fact about minimal polynomials mentioned above. So it is certainly arguable which of these facts is "more basic". My reason for choosing the latter is because I think it is a little more concrete and amenable to computation: indeed, if you don't know this fact about minimal polynomials, it seems to me that you're going to have a hard time computing any nontrivial examples of integral closures whatsoever.