Answer
$${f_x}\left( {x,y} \right) = \frac{{{y^{ - 1/2}}}}{{{y^2} + {x^2}}},\,\,\,\,\,\,\,\,\,{f_y}\left( {x,y} \right) = \frac{{x{y^{ - 3/2}}}}{{{y^2} + {x^2}}} - \frac{3}{2}{y^{ - 5/2}}{\tan ^{ - 1}}\left( {x/y} \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {y^{ - 3/2}}{\tan ^{ - 1}}\left( {x/y} \right) \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr
& {\text{use the quotient rule}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{y^{ - 3/2}}{{\tan }^{ - 1}}\left( {x/y} \right)} \right] \cr
& {f_x}\left( {x,y} \right) = {y^{ - 3/2}}\frac{\partial }{{\partial x}}\left[ {{{\tan }^{ - 1}}\left( {x/y} \right)} \right] \cr
& {\text{use the differentitation rule for }}\tan u \cr
& {f_x}\left( {x,y} \right) = {y^{ - 3/2}}\left( {\frac{{1/y}}{{1 + {{\left( {x/y} \right)}^2}}}} \right) \cr
& {\text{simplifying}} \cr
& {f_x}\left( {x,y} \right) = {y^{ - 5/2}}\left( {\frac{{{y^2}}}{{{y^2} + {x^2}}}} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{{{y^{ - 1/2}}}}{{{y^2} + {x^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
& {\text{use the quotient rule}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{y^{ - 3/2}}{{\tan }^{ - 1}}\left( {x/y} \right)} \right] \cr
& {\text{by the product rule}} \cr
& {f_y}\left( {x,y} \right) = {y^{ - 3/2}}\frac{\partial }{{\partial y}}\left[ {{{\tan }^{ - 1}}\left( {x/y} \right)} \right] + {\tan ^{ - 1}}\left( {x/y} \right)\frac{\partial }{{\partial y}}\left[ {{y^{ - 3/2}}} \right] \cr
& {\text{use the differentitation rule for }}\tan u \cr
& {f_y}\left( {x,y} \right) = {y^{ - 3/2}}\left( {\frac{{ - x/{y^2}}}{{1 + {{\left( {x/y} \right)}^2}}}} \right) + {\tan ^{ - 1}}\left( {x/y} \right)\left( { - \frac{3}{2}{y^{ - 5/2}}} \right) \cr
& {\text{simplifying}} \cr
& {f_y}\left( {x,y} \right) = \frac{{{y^{ - 7/2}}}}{x}\left( {\frac{{{y^2}}}{{{y^2} + {x^2}}}} \right) - \frac{3}{2}{y^{ - 5/2}}{\tan ^{ - 1}}\left( {x/y} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{{x{y^{ - 3/2}}}}{{{y^2} + {x^2}}} - \frac{3}{2}{y^{ - 5/2}}{\tan ^{ - 1}}\left( {x/y} \right) \cr} $$