Answer
$${g_u}\left( {u,v} \right) = \cos v - v\cos u\,\,\,\,\,\,\,\,\,\,\,{g_v}\left( {u,v} \right) = - u\sin v - \sin u$$
Work Step by Step
$$\eqalign{
& g\left( {u,v} \right) = u\cos v - v\sin u \cr
& {\text{Find the partial derivatives }}{g_u}\left( {u,v} \right){\text{ and }}{g_v}\left( {u,v} \right){\text{ then}} \cr
& {g_u}\left( {u,v} \right) = \frac{\partial }{{\partial u}}\left[ {u\cos v - v\sin u} \right] \cr
& {\text{treat }}v{\text{ as a constant}} \cr
& {g_u}\left( {u,v} \right) = \cos v\frac{\partial }{{\partial u}}\left[ u \right] - v\frac{\partial }{{\partial u}}\left[ {\sin u} \right] \cr
& {g_u}\left( {u,v} \right) = \cos v\left( 1 \right) - v\left( {\cos u} \right) \cr
& {g_u}\left( {u,v} \right) = \cos v - v\cos u \cr
& \cr
& and \cr
& \cr
& {g_v}\left( {u,v} \right) = \frac{\partial }{{\partial u}}\left[ {u\cos v - v\sin u} \right] \cr
& {\text{treat }}u{\text{ as a constant}} \cr
& {g_v}\left( {u,v} \right) = u\frac{\partial }{{\partial v}}\left[ {\cos v} \right] - \sin u\frac{\partial }{{\partial v}}\left[ v \right] \cr
& {g_v}\left( {u,v} \right) = u\left( { - \sin v} \right) - \sin u\left( 1 \right) \cr
& {g_v}\left( {u,v} \right) = - u\sin v - \sin u \cr} $$
