Solution of $\frac{xdx - ydy}{xdy - ydx} = \sqrt{\frac{1 - y^{2} + x^{2}}{x^{2} - y^{2}}}$ 
Find the solution of $$\frac{xdx - ydy}{xdy - ydx} = \sqrt{\frac{1 - y^{2} + x^{2}}{x^{2} - y^{2}}}$$

I was able to bring it down to $$\frac{d(x^2-y^2)}{\sqrt{1+x^2-y^2}}=2\left(\frac{x.d(y/x)}{\sqrt{1-(y/x)^2}}\right)$$
Any help would be greatly appreciated. Thanks in advance!
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Edit: An answer exists here, but I'm trying to solve the question $without$ using trigonometric or hyperbolic substitution (I was told it could be done without both, and that the solution was quite neat). 
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EDIT: Also the question that is mentioned as a possible duplicate is $\textbf{different}$ from mine.

We can write the equation as follows:
$$ \frac{xdx-ydy}{x^2-y^2}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}\frac{xdy-ydx}{x^2-y^2}$$
It's 
$$ \frac{xdx-ydy}{x^2-y^2}=\frac{d\sqrt{x^2-y^2}}{\sqrt{x^2-y^2}}$$
and
$$\frac{xdy-ydx}{x^2-y^2}=\frac{1}{2}d\ln\frac{x+y}{x-y}$$
$\,\displaystyle \sqrt{x^2-y^2}\,$ and $\,\displaystyle \ln\frac{x+y}{x-y}\,$ used as variables gives the wished result. 
