Answer
$$\frac{{\partial z}}{{\partial p}} = - \frac{{2q}}{{{{\left( {p - q} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}z = \frac{x}{y},\,\,\,\,x = p + q,\,\,\,\,{\text{and }}y = p - q \cr
& {\text{Write }}z{\text{ in terms of }}p{\text{ and }}q \cr
& z = \frac{{p + q}}{{p - q}} \cr
& {\text{Find }}\frac{{\partial z}}{{\partial p}} \cr
& \frac{{\partial z}}{{\partial p}} = \frac{{\left( {p - q} \right)\frac{\partial }{{\partial p}}\left[ {p + q} \right] - \left( {p + q} \right)\frac{\partial }{{\partial p}}\left[ {p - q} \right]}}{{{{\left( {p - q} \right)}^2}}} \cr
& \frac{{\partial z}}{{\partial p}} = \frac{{\left( {p - q} \right)\left( 1 \right) - \left( {p + q} \right)\left( 1 \right)}}{{{{\left( {p - q} \right)}^2}}} \cr
& \frac{{\partial z}}{{\partial p}} = \frac{{p - q - p - q}}{{{{\left( {p - q} \right)}^2}}} \cr
& \frac{{\partial z}}{{\partial p}} = - \frac{{2q}}{{{{\left( {p - q} \right)}^2}}} \cr} $$
