What I have so far is
Suppose $f: 3 \mathbb Z \rightarrow 9 \mathbb Z$ is a ring isomorphism.
Let $f(3)=9a $
Now let us compute $f(9)$ two different ways:
$$f(9)= f(3+3+3)=f(3)+f(3)+f(3)=9a+9a+9a=27a$$
and
$$f(9)= f(3\times 3)=f(3)\times f(3)=9a\times 9a=81a^2 $$
Thus we have $27a=81a^2$
So we have two integer solutionns of $a=1/3$ and $a=0$ which implies $f: 3 \mathbb Z \rightarrow 9 \mathbb Z$ is not an isomorphism.
I'm not completely sure if this proves it or if its even right. Any help would be greatly appreciated, thanks in advance!