Answer
$${f_x}\left( {x,y} \right) = - \frac{{4{{\sec }^2}x}}{{3{y^{8/3}}{{\left( {\tan x} \right)}^{7/3}}}},\,\,\,\,\,\,{f_y}\left( {x,y} \right) = - \frac{8}{{3{{\left( {\tan x} \right)}^{4/3}}{y^{11/3}}}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {\left( {{y^2}\tan x} \right)^{ - 4/3}} \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}{\left( {{y^2}\tan x} \right)^{ - 4/3}} \cr
& {f_x}\left( {x,y} \right) = {y^{ - 8/3}}\frac{\partial }{{\partial x}}{\left( {\tan x} \right)^{ - 4/3}} \cr
& {\text{By the chain rule}} \cr
& {f_x}\left( {x,y} \right) = {y^{ - 8/3}}\left( { - \frac{4}{3}} \right){\left( {\tan x} \right)^{ - 7/3}}\left( {{{\sec }^2}x} \right) \cr
& {f_x}\left( {x,y} \right) = - \frac{{4{{\sec }^2}x}}{{3{y^{8/3}}{{\left( {\tan x} \right)}^{7/3}}}} \cr
& \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ differentiating with respect to }}y;{\text{ treat }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}{\left( {{y^2}\tan x} \right)^{ - 4/3}} \cr
& {f_y}\left( {x,y} \right) = {\left( {\tan x} \right)^{ - 4/3}}\frac{\partial }{{\partial y}}\left( {{y^{ - 8/3}}} \right) \cr
& {\text{By the power rule}} \cr
& {f_y}\left( {x,y} \right) = {\left( {\tan x} \right)^{ - 4/3}}\frac{\partial }{{\partial x}}\left( {{y^{ - 8/3}}} \right) \cr
& {f_y}\left( {x,y} \right) = {\left( {\tan x} \right)^{ - 4/3}}\left( { - \frac{8}{3}{y^{ - 11/3}}} \right) \cr
& {f_y}\left( {x,y} \right) = - \frac{8}{{3{{\left( {\tan x} \right)}^{4/3}}{y^{11/3}}}} \cr} $$
