If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$

Both $x,y\in[-1,1]$. So, $x=\sin\theta$ and $y=\sin\phi$ for some $\displaystyle \theta,\phi\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$. Note that $\cos\theta, \cos\phi\ge0$.

We have $$\cos\theta+\cos\phi=a(\sin\theta-\sin\phi)$$

So, $$2\cos\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}=2a\cos\frac{\theta+\phi}{2}\sin\frac{\theta-\phi}{2}$$

$\displaystyle \tan\frac{\theta-\phi}{2}=\frac{1}{a}$ is a constant. So, $\displaystyle \frac{d\phi}{d\theta}=1$.

$$\frac{dy}{dx}=\frac{\cos\phi}{\cos\theta}\frac{d\phi}{d\theta}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$$

After my comment you will get $$a\sqrt{1-x^2}+x=\frac{1-xy+\sqrt{1-x^2}\sqrt{1-y^2}}{x-y}$$ and $$a\sqrt{1-y^2}-y=\frac{1-xy+\sqrt{1-x^2}\sqrt{1-y^2}}{x-y}$$ and you will get the desired result!

HINT : There is no parameter $a$ in the formula to be proved. So, first transform the initial equation into an equation where $a$ will be immediately eliminated by differentiation: $$\frac{\sqrt{1-x^2}+\sqrt{1-y^2}}{x-y}=a$$ Differentiate and simplify.