Let $A$ be a commutative ring and $p$, $q$ prime ideals such that $q \subseteq p$. Show that $A_p \otimes A/q = \dfrac{A_p}{qA_p}$. Here's what I have done -
Let $S = A \setminus p$. Then $A_p = S^{-1}A$.
So, $A_p \otimes A/q = S^{-1}A \otimes A/q \cong S^{-1}(A/q) \cong \dfrac{S^{-1}A}{S^{-1}q} = \dfrac{A_p}{qA_p}$.
Is this correct? I can't tell where $q \subseteq p$ is used.
As it was pointed out above, it is true in general that if you have an $R$-module $M$ and $I \subset R$ is an ideal then $$M \otimes R/I \cong M/IM.$$ It is easily derived from the fact that tensor products are right exact, meaning that tensoring s.e.s. $$0 \to I \to R \to R/I \to 0$$ with $M$ gives you an exact $$I \otimes M \overset{\alpha}{\to} R \otimes M \overset{\beta}{\to} R/I \otimes M \to 0.$$ Since $\beta$ is surjective, we get that $R/I \otimes M \cong R \otimes M /\ker \beta = R \otimes M / \operatorname{im} \alpha$. The only thing left to notice is that $R \otimes M$ is canonically isomorphic to $M$ and $\operatorname{im} \alpha $ becomes $IM$ under this isomorphism.
In your case, as was also pointed out above, the condition $q \subseteq p$ ensures that $q A_p \neq A_p$, as if $\exists x \in q \setminus p$ then $1 = \frac{x}{1} \cdot \frac{1}{x} \in q A_p$.