Given $$\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$$

$$ \Rightarrow $$ $${{dy} \over {dx}} = - {{\sqrt {1 - {y^2}} } \over {\sqrt {1 - {x^2}} }}$$

$$ \Rightarrow $$ $${{dy} \over {\sqrt {1 - {y^2}} }} + {{dx} \over {\sqrt {1 - {x^2}} }} = 0$$

$$ \Rightarrow $$ sin^-1 y + sin^-1 x = c

Given that $$y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}$$

So when x = $${1 \over 2}$$ then y = $${{\sqrt 3 } \over 2}$$

$$ \therefore $$ sin^-1 $${{\sqrt 3 } \over 2}$$ + sin^-1 $${1 \over 2}$$ = c

$$ \Rightarrow $$ c = $${\pi \over 3}$$ + $${\pi \over 6}$$ = $${\pi \over 2}$$

So sin^-1 y + sin^-1 x = $${\pi \over 2}$$

$$ \Rightarrow $$ sin^-1 y = $${\pi \over 2}$$ - sin^-1 x

$$ \Rightarrow $$ sin^-1 y = cos^-1 x

Now $$y\left( { - {1 \over {\sqrt 2 }}} \right)$$ means x = $$ - {1 \over {\sqrt 2 }}$$ and find y.

Putting x = $$ - {1 \over {\sqrt 2 }}$$

sin^-1 y = cos^-1 $$\left( { - {1 \over {\sqrt 2 }}} \right)$$ = $${{3\pi } \over 4}$$

y (at x = $$ - {1 \over {\sqrt 2 }}$$) = sin($${{3\pi } \over 4}$$) = $${{1 \over {\sqrt 2 }}}$$