Answer
$$y = - x + 1$$
Work Step by Step
$$\eqalign{
& \arctan \left( {x + y} \right) = {y^2} + \frac{\pi }{4},{\text{ }}\left( {1,0} \right) \cr
& {\text{Using implicit differentiation}} \cr
& \frac{d}{{dx}}\left[ {\arctan \left( {x + y} \right)} \right] = \frac{d}{{dx}}\left[ {{y^2} + \frac{\pi }{4}} \right] \cr
& \frac{{\frac{d}{{dx}}\left[ {x + y} \right]}}{{1 + {{\left( {x + y} \right)}^2}}} = \frac{d}{{dx}}\left[ {{y^2} + \frac{\pi }{4}} \right] \cr
& \frac{{1 + y'}}{{1 + {{\left( {x + y} \right)}^2}}} = 2yy' \cr
& {\text{Solve for }}y' \cr
& \frac{1}{{1 + {{\left( {x + y} \right)}^2}}} + \frac{{y'}}{{1 + {{\left( {x + y} \right)}^2}}} = 2yy' \cr
& \frac{{y'}}{{1 + {{\left( {x + y} \right)}^2}}} - 2yy' = - \frac{1}{{1 + {{\left( {x + y} \right)}^2}}} \cr
& \left[ {\frac{1}{{1 + {{\left( {x + y} \right)}^2}}} - 2y} \right]y' = - \frac{1}{{1 + {{\left( {x + y} \right)}^2}}} \cr
& {\text{Multiply both sides by }}1 + {\left( {x + y} \right)^2} \cr
& \left( {1 - 2y\left[ {1 + {{\left( {x + y} \right)}^2}} \right]} \right)y' = - 1 \cr
& y' = \frac{1}{{2y\left[ {1 + {{\left( {x + y} \right)}^2}} \right] - 1}} \cr
& {\text{Calculate }}y'{\text{ at the given point }}\left( {1,0} \right) \cr
& y' = \frac{1}{{2\left( 0 \right)\left[ {1 + {{\left( {1 + 0} \right)}^2}} \right] - 1}} \cr
& y' = - 1 \cr
& {\text{The slope at the given point is }}m = - 1 \cr
& {\text{The equation of the tangent line at }}\left( {1,0} \right){\text{ is}} \cr
& y - 0 = - \left( {x - 1} \right) \cr
& y = - x + 1 \cr} $$
