Differentiating $x^2=\frac{x+y}{x-y}$ Differentiate:
$$x^2=\frac{x+y}{x-y}$$
Preferring to avoid the quotient rule, I take away the fraction:
$$x^2=(x+y)(x-y)^{-1}$$
Then:
$$2x=(1+y')(x-y)^{-1}-(1-y')(x+y)(x-y)^{-2}$$
If I were to multiply the entire equation by $(x-y)^2$ then continue, I get the solution. However, if I continue the following, I don't. Likely some place I erred, but I cannot figure out where:
Expansion:
$$2x=(x-y)^{-1}+y'(x-y)^{-1}-(x+y)(x-y)^{-2}+y'(x+y)(x-y)^{-2}$$
Preparing to isolate for $y'$:
$$2x-(x-y)^{-1}+(x+y)(x-y)^{-2}=y'(x-y)^{-1}+y'(x+y)(x-y)^{-2}$$
$$2x-(x-y)^{-1}+(x+y)(x-y)^{-2}=y'[(x-y)^{-1}+(x+y)(x-y)^{-2}]$$
Isolating $y'$:
$$y'=\frac{2x-(x-y)^{-1}+(x+y)(x-y)^{-2}}{(x-y)^{-1}+(x+y)(x-y)^{-2}}$$
Multiple top and bottom by $(x-y)$:
$$y'=\frac{2x(x-y)-1+(x+y)(x-y)^{-1}}{1+(x+y)(x-y)^{-1}}$$
Then, inserting $x^2$ into $(x+y)(x-y)^{-1}$, I get:
$$y'=\frac{2x(x-y)-1+x^2}{1+x^2}$$
While the answer states:
$$y'=\frac{x(x-y)^2+y}{x}$$
Which I do get if I multiplied the entire equation by $(x-y)^2$ before. It does not seem to be another form of the answer, as putting $x=2$, the denominator cannot match each other. Where have I gone wrong?

You should first find $y$ in terms of $x$ as follows $$x^2=\frac{x+y}{x-y}$$ $$x^3-x^2y=x+y$$$$ (1+x^2)y=x^3-x$$
$$y=\frac{x^3-x}{1+x^2}$$
now, differentiate w.r.t. $x$, $$\frac{dy}{dx}=\frac{(1+x^2)(3x^2-1)-(x^3-x)(2x)}{(1+x^2)^2}$$
$$=\frac{x^4+4x^2-1}{(1+x^2)^2}$$
