Answer
$${f_x}\left( {x,y} \right) = 2y{x^{2y - 1}}\,\,\,\,\,\,\,\,\,{f_y}\left( {x,y} \right) = {x^{2y}}\ln \left( {{x^2}} \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {x^{2y}} \cr
& {\text{Find the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right){\text{ then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^{2y}}} \right] \cr
& {\text{treat }}y{\text{ as a constant}}{\text{, then it is possible use the power rule}} \cr
& {f_x}\left( {x,y} \right) = 2y{x^{2y - 1}} \cr
& \cr
& and \cr
& \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^{2y}}} \right] \cr
& {\text{treat }}x{\text{ as a constant}}{\text{, then }} \cr
& {f_y}\left( {x,y} \right) = {x^{2y}}\ln \left( x \right)\frac{\partial }{{\partial y}}\left[ {2y} \right] \cr
& {f_y}\left( {x,y} \right) = {x^{2y}}\ln \left( x \right)\left( 2 \right) \cr
& {f_y}\left( {x,y} \right) = 2{x^{2y}}\ln \left( x \right) \cr
& {f_y}\left( {x,y} \right) = {x^{2y}}\ln \left( {{x^2}} \right) \cr} $$
