Answer
$${f_x}\left( {x,y} \right) = - \frac{{{y^2} - 2xy}}{{{{\left( {x{y^2} - {x^2}y} \right)}^2}}}{\text{ and }}{f_y}\left( {x,y} \right) = - \frac{{2xy - {x^2}}}{{{{\left( {x{y^2} - {x^2}y} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \frac{1}{{x{y^2} - {x^2}y}} \cr
& {\text{Write the function as }} \cr
& f\left( {x,y} \right) = {\left( {x{y^2} - {x^2}y} \right)^{ - 1}} \cr
& \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{{\left( {x{y^2} - {x^2}y} \right)}^{ - 1}}} \right] \cr
& {f_x}\left( {x,y} \right) = - {\left( {x{y^2} - {x^2}y} \right)^{ - 2}}\frac{\partial }{{\partial x}}\left[ {x{y^2} - {x^2}y} \right] \cr
& {f_x}\left( {x,y} \right) = - {\left( {x{y^2} - {x^2}y} \right)^{ - 2}}\left( {{y^2} - 2xy} \right) \cr
& {f_x}\left( {x,y} \right) = - \frac{{{y^2} - 2xy}}{{{{\left( {x{y^2} - {x^2}y} \right)}^2}}} \cr
& \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{, differentiating with respect to }}y;{\text{ treat }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{{\left( {x{y^2} - {x^2}y} \right)}^{ - 1}}} \right] \cr
& {f_y}\left( {x,y} \right) = - {\left( {x{y^2} - {x^2}y} \right)^{ - 2}}\frac{\partial }{{\partial y}}\left[ {x{y^2} - {x^2}y} \right] \cr
& {f_y}\left( {x,y} \right) = - {\left( {x{y^2} - {x^2}y} \right)^{ - 2}}\left( {2xy - {x^2}} \right) \cr
& {f_y}\left( {x,y} \right) = - \frac{{2xy - {x^2}}}{{{{\left( {x{y^2} - {x^2}y} \right)}^2}}} \cr} $$
