Since $R[x]/(I)=(R/I)[x]$, and by the Chinese Remainder Theorem, $$\mathbb{Z}_6[x]/(3,x) \cong \mathbb{Z}_3[x]/(3,x) \times \mathbb{Z}_2[x]/(3,x) \cong \mathbb{Z}_3[x]/(f(x)) \times \mathbb{Z}_2[x]/(g(x)) \cong \mathbb{Z}_3[x] \times \mathbb{Z}_2[x]/(g(x))$$ since $\mathbb{Z}_2$, $\mathbb{Z}_3$ are fields...
is what I have so far. But I feel like it's trash. Can anyone help me where to begin?
To show that an ideal is principal, you should generally proceed by constructing a generator.
In this case, we can guess that the generator is of the form $a+bx$. By reducing modulo $2$ and $3$, we can solve for $a$ and $b$: $a+bx$ must generate the ideal $(x)$ in $\mathbb{Z}/3\mathbb{Z}[x]$ and the ideal $(1)$ in $\mathbb{Z}/2\mathbb{Z}[x]$.
Having done this, we have to show that $a+bx \in (3,x)$, and that $3,x\in (a+bx)$.