Answer
$${f_x}\left( {x,y} \right) = \frac{{4x}}{{{{\left( {4{x^2} + {y^2}} \right)}^{1/2}}}}{\text{ and }}{f_y}\left( {x,y} \right) = \frac{y}{{{{\left( {4{x^2} + {y^2}} \right)}^{1/2}}}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \sqrt {4{x^2} + {y^2}} \cr
& f\left( {x,y} \right) = {\left( {4{x^2} + {y^2}} \right)^{1/2}} \cr
& {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{{\left( {4{x^2} + {y^2}} \right)}^{1/2}}} \right] \cr
& {\text{use chain rule}} \cr
& {f_x}\left( {x,y} \right) = \frac{1}{2}{\left( {4{x^2} + {y^2}} \right)^{ - 1/2}}\frac{\partial }{{\partial x}}\left[ {4{x^2} + {y^2}} \right] \cr
& {f_x}\left( {x,y} \right) = \frac{1}{2}{\left( {4{x^2} + {y^2}} \right)^{ - 1/2}}\left( {8x} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{{4x}}{{{{\left( {4{x^2} + {y^2}} \right)}^{1/2}}}} \cr
& \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{{\left( {4{x^2} + {y^2}} \right)}^{1/2}}} \right] \cr
& {\text{treat x as a constant and y as a variable}}{\text{. then}} \cr
& {f_y}\left( {x,y} \right) = \frac{1}{2}{\left( {4{x^2} + {y^2}} \right)^{ - 1/2}}\frac{\partial }{{\partial y}}\left[ {4{x^2} + {y^2}} \right] \cr
& {\text{use chain rule}} \cr
& {f_y}\left( {x,y} \right) = \frac{1}{2}{\left( {4{x^2} + {y^2}} \right)^{ - 1/2}}\left( {2y} \right) \cr
& {f_y}\left( {x,y} \right) = y{\left( {4{x^2} + {y^2}} \right)^{ - 1/2}} \cr
& {f_y}\left( {x,y} \right) = \frac{y}{{{{\left( {4{x^2} + {y^2}} \right)}^{1/2}}}} \cr} $$
