Answer
$${f_x}\left( {x,y} \right) = - \frac{{2x}}{{1 + {{\left( {{x^2} + {y^2}} \right)}^2}}}{\text{ and }}{f_y}\left( {x,y} \right) = - \frac{{2y}}{{1 + {{\left( {{x^2} + {y^2}} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 1 - {\tan ^{ - 1}}\left( {{x^2} + {y^2}} \right) \cr
& {\text{Find the first partial derivative }}{f_x}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {1 - {{\tan }^{ - 1}}\left( {{x^2} + {y^2}} \right)} \right] \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ 1 \right] - \frac{\partial }{{\partial x}}\left[ {{{\tan }^{ - 1}}\left( {{x^2} + {y^2}} \right)} \right] \cr
& {\text{treat }}y{\text{ as a constant}}{\text{, then use }}\left( {{{\tan }^{ - 1}}u} \right)' = \frac{1}{{1 + {u^2}}}\left( {u'} \right) \cr
& {f_x}\left( {x,y} \right) = 0 - \frac{1}{{1 + {{\left( {{x^2} + {y^2}} \right)}^2}}}\frac{\partial }{{\partial x}}\left[ {{x^2} + {y^2}} \right] \cr
& {f_x}\left( {x,y} \right) = - \frac{1}{{1 + {{\left( {{x^2} + {y^2}} \right)}^2}}}\left( {2x} \right) \cr
& {f_x}\left( {x,y} \right) = - \frac{{2x}}{{1 + {{\left( {{x^2} + {y^2}} \right)}^2}}} \cr
& \cr
& {\text{Find the first partial derivative }}{f_y}\left( {x,y} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {1 - {{\tan }^{ - 1}}\left( {{x^2} + {y^2}} \right)} \right] \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ 1 \right] - \frac{\partial }{{\partial y}}\left[ {{{\tan }^{ - 1}}\left( {{x^2} + {y^2}} \right)} \right] \cr
& {\text{treat }}x{\text{ as a constant}}{\text{, then use }}\left( {{{\tan }^{ - 1}}u} \right)' = \frac{1}{{1 + {u^2}}}\left( {u'} \right) \cr
& {f_y}\left( {x,y} \right) = 0 - \frac{1}{{1 + {{\left( {{x^2} + {y^2}} \right)}^2}}}\frac{\partial }{{\partial y}}\left[ {{x^2} + {y^2}} \right] \cr
& {f_y}\left( {x,y} \right) = - \frac{1}{{1 + {{\left( {{x^2} + {y^2}} \right)}^2}}}\left( {2y} \right) \cr
& {f_y}\left( {x,y} \right) = - \frac{{2y}}{{1 + {{\left( {{x^2} + {y^2}} \right)}^2}}} \cr} $$
