Problem is ring isomorphism from $\mathbb Z_3\times \mathbb Z_2$ to $\mathbb Z_6$.
Let $f=\mathbb Z_3\times \mathbb Z_2 \to \mathbb Z_6$ defined as $f(a_3,b_2)=ab_6$.
I prove it is isomorphism:
It is obviously that $|\mathbb Z_6|$=$|\mathbb Z_3\times\mathbb Z_2 |$ (the same cardinality). Surjectivity is obviously because of the Chinese Remainder Theorem.
Injectivity : Let $f(a_3,b_2)=f(c_3,d_2)$. Then $f(a_3-c_3,b_2-d_2)=f(a_3,b_2)-f(c_3,d_2)=0$. If $f(a_3,b_2)=ab_6=0_6$ then because a_2 is divisible by 2 and b_3 by 3, then ab_6 is divisible by 6. Therefore ab_6=0.
It is correct ? Thanks