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