Answer
$${f_x}\left( {x,y} \right) = \frac{{2{y^2} - 6{x^2} - 30x{y^2}}}{{{{\left( {3{x^2} + {y^2}} \right)}^2}}}{\text{ and }}{f_y}\left( {x,y} \right) = \frac{{30{x^2}y - 4xy}}{{{{\left( {3{x^2} + {y^2}} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \frac{{2x + 5{y^2}}}{{3{x^2} + {y^2}}} \cr
& {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{2x + 5{y^2}}}{{3{x^2} + {y^2}}}} \right] \cr
& {\text{by using the quotient rule}} \cr
& {f_x}\left( {x,y} \right) = \frac{{\left( {3{x^2} + {y^2}} \right)\frac{\partial }{{\partial x}}\left[ {2x + 5{y^2}} \right] - \left( {2x + 5{y^2}} \right)\frac{\partial }{{\partial x}}\left[ {3{x^2} + {y^2}} \right]}}{{{{\left( {3{x^2} + {y^2}} \right)}^2}}} \cr
& {\text{solving derivatives}} \cr
& {f_x}\left( {x,y} \right) = \frac{{\left( {3{x^2} + {y^2}} \right)\left( 2 \right) - \left( {2x + 5{y^2}} \right)\left( {6x} \right)}}{{{{\left( {3{x^2} + {y^2}} \right)}^2}}} \cr
& {f_x}\left( {x,y} \right) = \frac{{6{x^2} + 2{y^2} - 12{x^2} - 30x{y^2}}}{{{{\left( {3{x^2} + {y^2}} \right)}^2}}} \cr
& {f_x}\left( {x,y} \right) = \frac{{2{y^2} - 6{x^2} - 30x{y^2}}}{{{{\left( {3{x^2} + {y^2}} \right)}^2}}} \cr
& \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{2x + 5{y^2}}}{{3{x^2} + {y^2}}}} \right] \cr
& {\text{by using the quotient rule}} \cr
& {f_y}\left( {x,y} \right) = \frac{{\left( {3{x^2} + {y^2}} \right)\frac{\partial }{{\partial y}}\left[ {2x + 5{y^2}} \right] - \left( {2x + 5{y^2}} \right)\frac{\partial }{{\partial y}}\left[ {3{x^2} + {y^2}} \right]}}{{{{\left( {3{x^2} + {y^2}} \right)}^2}}} \cr
& {\text{solving derivatives}} \cr
& {f_y}\left( {x,y} \right) = \frac{{\left( {3{x^2} + {y^2}} \right)\left( {10y} \right) - \left( {2x + 5{y^2}} \right)\left( {2y} \right)}}{{{{\left( {3{x^2} + {y^2}} \right)}^2}}} \cr
& {f_y}\left( {x,y} \right) = \frac{{30{x^2}y + 10{y^3} - 4xy - 10{y^3}}}{{{{\left( {3{x^2} + {y^2}} \right)}^2}}} \cr
& {f_y}\left( {x,y} \right) = \frac{{30{x^2}y - 4xy}}{{{{\left( {3{x^2} + {y^2}} \right)}^2}}} \cr} $$
