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