Problem
Show that $A={(3x,y)|x,y \in Z}$ is a maximal ideal . Generalize the result.
Proof
$A$ is an ideal
For any$ (a,b) \in Z+Z$ and $(3x,y) \in A, (a,b)(3x,y) \in Z$. Also it is easy to see $A$ is a subring.
$A$ is a maximal ideal
Consider a maximal ideal $B$ which contains $A$. Now$ (c,d) \in B$ such that $gcd(3,c)=1. 1=3\alpha+c\beta$. Then $(1,1)=(3\alpha+c\beta,1-d+d)$ . Hence $(1,1) \in B$. So $A$ is maximal.
Doubt
1.What is the generalization of the result?
$A$={$(nx,y)|x,y \in Z$}
2.Is my proof ok?
With respect to your proof, the idea is correct (in fact I find it quite original), but to be honest, it is terribly written. Let me rewrite it appropriately:
As you have said, it is easy to check that $A$ is indeed an ideal (another way to argue is to observe that $A=3\mathbb{Z} \times \mathbb{Z}$ is a product of two ideals of $\mathbb{Z}$). Let $B \subset \mathbb{Z} \times \mathbb{Z}$ a maximal ideal containing $A$. Arguing by contradiction, we suppose that the inclusion is strict; that is, there exists $(c,d) \in B$ such that $\text{gcd}(3,c)=1$. By Bézout's Identity, there exist $\alpha, \beta \in \mathbb{Z}$ such that $$3\alpha+c\beta=1.$$ Hence, $$(1,1)=(3\alpha,1-d)+(\beta,1)(c,d) \in B,$$ against $B$ being proper. Therefore, $A=B$ and $A$ is indeed a maximal ideal.
We now come to consider the question of the generalization. For every $n \geq 0$ we denote $$A_n:=\{(nx,y) / x,y \in \mathbb{Z}\},$$ and observe that in fact $$A_n=n\mathbb{Z} \times \mathbb{Z}.$$ Therefore, $A_n$ is a product of ideals of $\mathbb{Z}$ and hence, an ideal of $\mathbb{Z} \times \mathbb{Z}$. Then: $$\frac{\mathbb{Z} \times \mathbb{Z}}{A_n}=\frac{\mathbb{Z} \times \mathbb{Z}}{n\mathbb{Z} \times \mathbb{Z}} \simeq \frac{\mathbb{Z}}{n\mathbb{Z}} \times \frac{\mathbb{Z}}{\mathbb{Z}} = \frac{\mathbb{Z}}{n\mathbb{Z}} \times \{0+\mathbb{Z}\} \simeq \frac{\mathbb{Z}}{n\mathbb{Z}}.$$ Thus, $$A_n \text{ is a maximal ideal} \iff \mathbb{Z} \times \mathbb{Z}/A_n \text{ is a field} \iff \mathbb{Z}/n\mathbb{Z} \text{ is a field} \iff n \text{ is a prime number.}$$ This also solves the original problem for $n=3$.