Let $A$ be a ring, $\mathfrak{q}$ an prime ideal and $\mathfrak{p}$ a maximal ideal such that $\mathfrak{q}\subset \mathfrak{p}$. Consider now the localized ring
$$(A/\mathfrak{q})_{\mathfrak{p}}$$
and denote by $\mathfrak{m}$ its maximal ideal. I saw somewhere that we have
$$\mathfrak{m}=(\mathfrak{p}+\mathfrak{q})/\mathfrak{q}$$
and so
$$\mathfrak{m}/\mathfrak{m}^2= \mathfrak{p}/(\mathfrak{p}^2+\mathfrak{q}).$$
I have two questions:
I dont understand how to get the expression of $\mathfrak{m}$ from the expression of the maximal ideal of a localization. So far I got something like $\mathfrak{m}=\mathfrak{p}(A_{\mathfrak{p}}/\mathfrak{q}_{\mathfrak{p}})$.
Assuming the expression of $\mathfrak{m}$, I dont understand the computation of the vector space $\mathfrak{m}/\mathfrak{m}^2$.
For the sake of completeness this is part of the proof of the Jacobian criterion for singularity of varieties.