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

My Attempt
$$
\frac{-2x}{2\sqrt{1-x^2}}-\frac{2y}{2\sqrt{1-y^2}}.\frac{dy}{dx}=a-a\frac{dy}{dx}\\
\implies \frac{dy}{dx}\bigg[a-\frac{y}{\sqrt{1-y^2}}\bigg]=a+\frac{x}{\sqrt{1-x^2}}\\
\frac{dy}{dx}=\frac{a\sqrt{1-x^2}+x}{\sqrt{1-x^2}}.\frac{\sqrt{1-y^2}}{a\sqrt{1-y^2}+x}=\sqrt{\frac{1-y^2}{1-x^2}}.\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}
$$
How do I poceed further and find the derivative ?

Let us remark that the looked for differential equation can be written under the form  
$$\frac{dx}{\sqrt{1-x^2}}=\frac{dy}{\sqrt{1-y^2}}\tag{1}$$
involving solutions of the form :
$$-\arccos(x)=-\arccos(y)+a\tag{2}$$
otherwise said with cartesian equation :
$$y=\cos(\arccos(x)-a)\tag{3}$$
where $a$ is any real.
Setting $\alpha=\arccos(x)$ in (3), we get the equivalent parametric equations :
$$\begin{cases}x&=&\cos(\alpha)\\y&=&\cos(\alpha-a)\end{cases}\tag{4}$$
in which we recognize that we are working with elliptical arcs, as shown on the graphical representation of the family of curves $C_a$ displayed below (we have to pay attention to the domains of variables $x$ and $y$ : in general we will not have the whole ellipses as solutions but arcs of them). 
From (4), it is easy to establish a connection with relationship:
$$\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$$
(in the spirit of the solution given by @CY Aries).

Fig. 1:Curves $C_a$ for $a=-\pi$ to $a=\pi$ with step $\pi/8$ (progressively changing from blue to red). The curves are elliptical arcs with two degenerate cases (straight lines).
