If $a$ is an arbitrary constant, then solution of the differential equation $\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0$, is

$x \sqrt{1-y^2}+y \sqrt{1-x^2}=a$

We have,

$\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0$

$\Rightarrow \sqrt{1-x^2} d y+\sqrt{1-y^2} d x=0$

$\Rightarrow \frac{1}{\sqrt{1-x^2}} d x+\frac{1}{\sqrt{1-y^2}} d y=0$

$\Rightarrow \sin ^{-1} x+\sin ^{-1} y=\sin ^{-1} a$           [On integrating]

$\Rightarrow \sin ^{-1}\left(x \sqrt{1-y^2}+y \sqrt{1-x^2}\right)=\sin ^{-1} a$

$\Rightarrow x \sqrt{1-y^2}+y \sqrt{1-x^2}=a$