Answer
$$\eqalign{
& {f_x}\left( {x,y} \right) = \frac{{15{x^4}y - 21{x^2}y}}{{2\sqrt {3{x^5} - 7{x^3}y} }} \cr
& {f_y}\left( {x,y} \right) = \frac{{3{x^5} - 7{x^3}}}{{2\sqrt {3{x^5} - 7{x^3}y} }} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \sqrt {3{x^5}y - 7{x^3}y} \cr
& \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\sqrt {3{x^5}y - 7{x^3}y} } \right] \cr
& {\text{Use the chain rule}} \cr
& {f_x}\left( {x,y} \right) = \frac{1}{2}{\left( {3{x^5}y - 7{x^3}y} \right)^{ - 1/2}}\frac{\partial }{{\partial x}}\left[ {3{x^5}y - 7{x^3}y} \right] \cr
& {f_x}\left( {x,y} \right) = \frac{1}{2}{\left( {3{x^5} - 7{x^3}y} \right)^{ - 1/2}}\left( {15{x^4}y - 21{x^2}y} \right) \cr
& {\text{simplifying}} \cr
& {f_x}\left( {x,y} \right) = \frac{{15{x^4}y - 21{x^2}y}}{{2{{\left( {3{x^5} - 7{x^3}y} \right)}^{1/2}}}} \cr
& {f_x}\left( {x,y} \right) = \frac{{15{x^4}y - 21{x^2}y}}{{2\sqrt {3{x^5} - 7{x^3}y} }} \cr
& \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ differentiating with respect to }}y;{\text{ treat }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\sqrt {3{x^5}y - 7{x^3}y} } \right] \cr
& {\text{Use the chain rule}} \cr
& {f_y}\left( {x,y} \right) = \frac{1}{2}{\left( {3{x^5}y - 7{x^3}y} \right)^{ - 1/2}}\frac{\partial }{{\partial y}}\left[ {3{x^5}y - 7{x^3}y} \right] \cr
& {f_y}\left( {x,y} \right) = \frac{1}{2}{\left( {3{x^5}y - 7{x^3}y} \right)^{ - 1/2}}\left( {3{x^5} - 7{x^3}} \right) \cr
& {\text{simplifying}} \cr
& {f_y}\left( {x,y} \right) = \frac{{3{x^5} - 7{x^3}}}{{2{{\left( {3{x^5} - 7{x^3}y} \right)}^{1/2}}}} \cr
& {f_y}\left( {x,y} \right) = \frac{{3{x^5} - 7{x^3}}}{{2\sqrt {3{x^5} - 7{x^3}y} }} \cr} $$
