Answer
$$y = - \frac{{2\pi }}{{\pi + 8}}x - \frac{{{\pi ^2}}}{{2\pi + 16}} + 1$$
Work Step by Step
$$\eqalign{
& {x^2} + x\arctan y = y - 1 \cr
& {\text{Using implicit differentiation}} \cr
& \frac{d}{{dx}}\left[ {{x^2} + x\arctan y} \right] = \frac{d}{{dx}}\left[ {y - 1} \right] \cr
& 2x + x\left( {\frac{1}{{1 + {y^2}}}} \right)y' + \arctan y = y' \cr
& {\text{Solve for }}y' \cr
& x\left( {\frac{1}{{1 + {y^2}}}} \right)y' - y' = - \arctan y - 2x \cr
& y'\left( {\frac{x}{{1 + {y^2}}} - 1} \right) = - \arctan y + 2x \cr
& y' = - \frac{{\arctan y + 2x}}{{\frac{x}{{1 + {y^2}}} - 1}} \cr
& {\text{Calculate }}y'{\text{ at the given point }}\left( { - \frac{\pi }{4},1} \right) \cr
& y' = \frac{{\arctan \left( 1 \right) + 2\left( { - \pi /4} \right)}}{{\frac{{ - \pi /4}}{{1 + {{\left( 1 \right)}^2}}} - 1}} \cr
& y' = - \frac{{2\pi }}{{\pi + 8}} \cr
& {\text{The slope at the given point is }}m = - \frac{{2\pi }}{{\pi + 8}} \cr
& {\text{The equation of the tangent line at }}\left( { - \frac{\pi }{4},1} \right){\text{ is}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 1 = - \frac{{2\pi }}{{\pi + 8}}\left( {x + \frac{\pi }{4}} \right) \cr
& y = - \frac{{2\pi }}{{\pi + 8}}x - \frac{{{\pi ^2}}}{{2\pi + 16}} + 1 \cr} $$
