Answer
$${f_x}\left( {x,y} \right) = \frac{{4x}}{{2{x^2} + {y^2}}}{\text{ and }}{f_y}\left( {x,y} \right) = \frac{{2y}}{{2{x^2} + {y^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \ln \left| {2{x^2} + {y^2}} \right| \cr
& {\text{find }}{f_x}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\ln \left( {2{x^2} + {y^2}} \right)} \right] \cr
& {\text{use the rule }}\left( {\ln u} \right)' = \frac{1}{u}\left( {u'} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{1}{{2{x^2} + {y^2}}}\frac{\partial }{{\partial x}}\left[ {2{x^2} + {y^2}} \right] \cr
& {\text{solve the derivative treating }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{1}{{2{x^2} + {y^2}}}\left( {4x} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{{4x}}{{2{x^2} + {y^2}}} \cr
& \cr
& {\text{find }}{f_y}\left( {x,y} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\ln \left( {2{x^2} + {y^2}} \right)} \right] \cr
& {\text{use the rule }}\left( {\ln u} \right)' = \frac{1}{u}\left( {u'} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{1}{{2{x^2} + {y^2}}}\frac{\partial }{{\partial y}}\left[ {2{x^2} + {y^2}} \right] \cr
& {\text{solve the derivative treating }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{1}{{2{x^2} + {y^2}}}\left( {2y} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{{2y}}{{2{x^2} + {y^2}}} \cr} $$
