Show that no ring containing R can contain a root of g(x) = 3x +1

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Show that if $R = \mathbb Z_6$ and $g(x) = 3x + 1 ∈ R[x]$, then $R[x]/(g(x)R[x])$ does not contain a root of $g(x)$. More generally, show that no ring containing $R$ can contain a root of $g(x)$.

$g(x)$ has one root $ \alpha =-1/3$ in $\mathbb R[x]$ But since $\alpha$ does not belong to $\mathbb Z_6=R$, there are no roots in $R[x]/(g(x)R[x])$
Is this correct? Also how do I prove the second part?

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Your solution for the first part is wrong. The existence of a root in a field does not imply that the polynomial has no roots in another field or ring. For example, consider the polynomial $4x+2$. Its root in $\Bbb R$ is $1/2$ and its roots in $\Bbb Z_6$ are $2$ and $-1$.

Hint for the second part: If $3x=-1$ then $0=6x=-2$.