partially differentiate $\frac{x-y}{\sqrt{x^2-y^2}}$ How do you partially differentiate $\frac{x-y}{\sqrt{x^2-y^2}}$, with respect to $y$ then $x$? 
I have done it and got the answer $$\frac{y^2+yx}{\sqrt[3]{x^2-y^2}}$$
However, the answer given in my paper gives it with a $-y^2$ on the top with all the rest the same. Any help?

We should get the following steps:
\begin{align}\frac{\partial}{\partial y}\left(\frac{x-y}{\sqrt{x^2-y^2}}\right)&=\frac{-x}{(x+y)\sqrt{x^2-y^2}}\\
\frac{\partial}{\partial x}\left(\frac{-x}{(x+y)\sqrt{x^2-y^2}}\right)&=\frac{x^2-xy+y^2}{(x-y)(x+y)^2\sqrt{x^2-y^2}}\end{align}
