<s>[INST] If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$? The question states:

If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$?

The options are:

  
*
  
*$$\sqrt{\frac{1-x^2}{1-y^2}}$$
  
*$$\sqrt{\frac{1-y^2}{1-x^2}}$$
  
*$$\frac{1-x^2}{1-y^2}$$
  
*$$\frac{1-y^2}{1-x^2}$$
  

I tried differentiating the entire expression:
$$\begin{align}
\frac{-2x}{2\sqrt{1-x^2}}+\frac{-2y}{2\sqrt{1-y^2}}\frac{dy}{dx} &= a(1-\frac{dy}{dx}) \\
a + \frac{x}{\sqrt{1-x^2}} &= \frac{dy}{dx}(a-\frac{y}{\sqrt{1-y^2}}) \\
\frac{dy}{dx} &= (a+\frac{x}{\sqrt{1-x^2}})(a-\frac{y}{\sqrt{1-y^2}})^{-1}
\end{align}$$
I am not sure how to proceed now. Is there a better approach? Is it advisable to use my approach?
 [/INST] Your approach is absolutely fine and the result that you have obtained is correct. If you handed me a homework like this I would be satisfied, since it is clear that you have understood the procedure of implicit differentiation.
Still, the problem is somewhat pedantic and asks you to do a final superfluous simplification: note that $a$ is absent from all of the available options, therefore express $a$ as $\dfrac {\sqrt{1-x^2} + \sqrt{1-y^2}} {x-y}$ and replace it in your own result. After a number of elementary algebraic simplifications you will obtain $\dfrac {\sqrt{1-y^2}} {\sqrt{1-x^2}}$, which is option 2.
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