Answer
$${f_x}\left( {x,y} \right) = - \frac{{2x{y^3}}}{{2 - {x^2}{y^3}}}{\text{ and }}{f_y}\left( {x,y} \right) = - \frac{{3{x^2}{y^2}}}{{2 - {x^2}{y^3}}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \ln \left| {2 - {x^2}{y^3}} \right| \cr
& {\text{find }}{f_x}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\ln \left| {2 - {x^2}{y^3}} \right|} \right] \cr
& {\text{use the rule }}\left( {\ln u} \right)' = \frac{1}{u}\left( {u'} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{1}{{2 - {x^2}{y^3}}}\frac{\partial }{{\partial x}}\left[ {2 - {x^2}{y^3}} \right] \cr
& {\text{solve the derivative treating }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{1}{{2 - {x^2}{y^3}}}\left( { - 2x{y^3}} \right) \cr
& {f_x}\left( {x,y} \right) = - \frac{{2x{y^3}}}{{2 - {x^2}{y^3}}} \cr
& \cr
& {\text{find }}{f_y}\left( {x,y} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\ln \left| {2 - {x^2}{y^3}} \right|} \right] \cr
& {\text{use the rule }}\left( {\ln u} \right)' = \frac{1}{u}\left( {u'} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{1}{{2 - {x^2}{y^3}}}\frac{\partial }{{\partial y}}\left[ {2 - {x^2}{y^3}} \right] \cr
& {\text{solve the derivative treating }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{1}{{2 - {x^2}{y^3}}}\left( { - 3{x^2}{y^2}} \right) \cr
& {f_y}\left( {x,y} \right) = - \frac{{3{x^2}{y^2}}}{{2 - {x^2}{y^3}}} \cr} $$
