Show that $\mathbb{Z}_{(3)}$ is not a field 
Given the localised ring $\mathbb{Z}_{(3)}=\{\frac{n}{3^k}|n \in \mathbb{Z} ;k\in\mathbb{N}\}$, I want to show that  this is not a field.

The main condition for a field is the existence of an inverse, such that:
$\exists y \in \mathbb{Z}_{(3)}:\frac{n}{3^k}\cdot y=1$
$\implies \frac{n}{3^k}\cdot \frac{m}{3^l}=1 $ (for a certain $m\in \mathbb{Z} ;l\in\mathbb{N}$)$\implies nm=3^{k+l}$, which is not always fullfilled, because of an counter example, $n=1$ and $m=2$. Is this a correct proof? Or do I have to go further?
 A: You have correctly identified what you need to do: you need to show that there exists $\frac{n}{3^k}\in\mathbb{Z}_{(3)}$ with the property that there does not exist an $\frac{m}{3^{\ell}}\in\mathbb{Z}_{(3)}$ such that $\frac{n}{3^k}\cdot \frac{m}{3^{\ell}} = 1$.
However, I do not think you have correctly done this. The obvious way to accomplish this is, again as you have tried to do, to simply exhibit a specific $n$ and a specific $k$ and show it has the desired property. That means showing that for this specific choice of $n$ and $k$, no choice of $m$ and $\ell$ will have the property that $\frac{n}{3^k}\cdot \frac{m}{3^{\ell}} = 1$. That suggests that your answer should:


*

*Specify an $n$ and a $k$; and

*Prove that no choice of $m$ and $\ell$ will work for that $n$ and $k$.


However, what you do instead is specify $n$ and $m$. 
Moreover, for your choice of $n$, namely $n=1$, it is not true that $\frac{n}{3^k}$ has no inverse! If $n=1$ and $k$ is an arbitrary nonnegative integer, then $\frac{n}{3^k} = \frac{1}{3^k}$ does have an inverse in $\mathbb{Z}_{(3)}$: namely,$\frac{3^{k+1}}{3}$ is an inverse for $\frac{1}{3^k}$. 
You are close, though: if you take $n=2$ and $k=1$, say, to get $\frac{2}{3}$, then you want to show that there is no choice of $m$ and $\ell$ that will satisy $\frac{2}{3}\cdot \frac{m}{3^{\ell}} = 1$. Do that. 
A: Your idea for a proof is fine, but the details are not. The simplest argument is that $2 \in \mathbb{Z}_{(3)}$ but its inverse $\frac12 \notin \mathbb{Z}_{(3)}$.
Here is a slightly more general fact, which solves your problem:
The smallest field that contains $\mathbb Z$ is $\mathbb Q$. Therefore, no intermediate ring $\mathbb Z \subsetneq R \subsetneq  \mathbb Q$ can be a field.
A: 
The main condition for a field is the existence of an inverse, such that:
  $\exists y \in \mathbb{Z}_{(3)}:\frac{n}{3^k}\cdot y=1$
$\implies \dfrac{n}{3^{\large k}}\, \dfrac{m}{3^{\large \ell}}=1\, $ (for a certain $m\in \mathbb{Z};\,\ell\in\mathbb{N})$ $\implies \color{#c00}{nm=3^{\large k+\ell}}$, which is not always fullfilled

hence $\,n/ 3^k$ invertible $\Rightarrow\, \color{#c00}{n\mid 3^{k+\ell}},\,$ so $\,n=2\,$ yields noninvertibles, e.g. $\,2\,$ is not invertible. $ $ QED

[...] counter example, $n=1$ and $m=2$. Is this a correct proof? Or do I have to go further?

No, $\,1/3^{\large k}$ has inverse $3^{\large k}$. Maybe you meant $\,n=2\,$ as above.  Going slightly further we can show from above that the only invertibles are $\,\pm 3^{\large k}\,$ for $\,k\in\Bbb Z$
