General solution of $\frac{dy}{dx}+\sqrt{\frac{1-y^2}{1-x^2}}=0$ is :

$sin^{-1}x+sin^{-1}y=c$

Given differential equation:

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

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

Separating variables:

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

Integrating both sides:

$\sin^{-1}y = -\sin^{-1}x + C$

final answer: The general solution is $\sin^{-1}y + \sin^{-1}x = C$