Answer
${w_u} = \frac{1}{{{v^2}}}{\text{ and }}{w_v} = - \frac{{2u}}{{{v^3}}}$
Work Step by Step
$$\eqalign{
& w = \frac{u}{{{v^2}}} \cr
& {\text{Find }}{w_u}{\text{ differentiate with respect to }}u \cr
& {w_u} = \frac{\partial }{{\partial u}}\left[ {\frac{u}{{{v^2}}}} \right] \cr
& {\text{Considering }}v{\text{ as a constant}} \cr
& {w_u} = \frac{1}{{{v^2}}}\frac{\partial }{{\partial u}}\left[ u \right] \cr
& {w_u} = \frac{1}{{{v^2}}}\left( 1 \right) \cr
& {w_u} = \frac{1}{{{v^2}}} \cr
& \cr
& {\text{Find }}{w_v}{\text{ differentiate with respect to }}v \cr
& {w_v} = \frac{\partial }{{\partial v}}\left[ {\frac{u}{{{v^2}}}} \right] \cr
& {\text{Considering }}u{\text{ as a constant}} \cr
& {w_v} = u\frac{\partial }{{\partial v}}\left[ {\frac{1}{{{v^2}}}} \right] \cr
& {w_v} = u\frac{\partial }{{\partial v}}\left[ {{v^{ - 2}}} \right] \cr
& {w_v} = u\left( { - 2{v^{ - 3}}} \right) \cr
& {w_v} = - \frac{{2u}}{{{v^3}}} \cr} $$
