Official solution
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(\[{{3\pi } \over 4}\]) = \[{{1 \over {\sqrt 2 }}}\]