Answer
$${f_x}\left( {x,y} \right) = \sqrt {{y^3}} \,\,\,\,\,\,\,\,\,\,\,\,{f_y}\left( {x,y} \right) = \frac{3}{2}x\sqrt y $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \sqrt {{x^2}{y^3}} \cr
& f\left( {x,y} \right) = \sqrt {{x^2}} \sqrt {{y^3}} \cr
& f\left( {x,y} \right) = x\sqrt {{y^3}} \cr
& \cr
& {\text{Find the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right){\text{ then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {x\sqrt {{y^3}} } \right] \cr
& {\text{treat }}y{\text{ as a constant}}{\text{,}} \cr
& {f_x}\left( {x,y} \right) = \sqrt {{y^3}} \frac{\partial }{{\partial x}}\left[ x \right] \cr
& {f_x}\left( {x,y} \right) = \sqrt {{y^3}} \left( 1 \right) \cr
& {f_x}\left( {x,y} \right) = \sqrt {{y^3}} \cr
& \cr
& and \cr
& \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {x\sqrt {{y^3}} } \right] \cr
& {\text{treat }}x{\text{ as a constant}}{\text{, then }} \cr
& {f_y}\left( {x,y} \right) = x\frac{\partial }{{\partial y}}\left[ {{y^{3/2}}} \right] \cr
& {f_y}\left( {x,y} \right) = x\left( {\frac{3}{2}{y^{1/2}}} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{3}{2}x{y^{1/2}} \cr
& {f_y}\left( {x,y} \right) = \frac{3}{2}x\sqrt y \cr} $$
