I am not able to go from the left hand side to the right hand side. Could someone take me through the steps, I tried using common denominator (y^2-x)^2, but was not able to get the result on the right hand side of the equation.
$\frac{2}{y^2-x}\left(-y \left(\frac{y-x^2}{y^2-x}\right)^2+\frac{y-x^2}{y^2-x}-x\right)=\frac{2}{y^2-x}\left(-\frac{2 \text{xy}}{\left(y^2-x\right)^2}\right)$
Leaving aside the $\displaystyle\frac{2}{y^2-x}$ wich is common to both sides of the identity:
$$-y\left(\frac{y-x^2}{y^2-x}\right)^2 + \frac{y-x^2}{y^2-x}- x = -y\left(\frac{y-x^2}{y^2-x}\right)^2 + \frac{y-x^2}{y^2-x}ยท\frac{y^2-x}{y^2-x}- x \left(\frac{y^2-x}{y^2-x}\right)^2 = \frac{-y(y-x^2)^2 + (y-x^2)(y^2-x) -x(y^2-x)^2}{(y^2-x)^2} = \\ \frac{-y^3+2x^2y^2-x^4y + y^3-xy-x^2y^2+ x^3 -xy^4+2x^2y^2-x^3}{(y^2-x)^2} = \\ \frac{3x^2y^2-x^4y -xy -xy^4}{(y^2-x)^2}= \frac{-xy(x^3 -3xy+y^3+1)}{(y^2-x)^2}$$
So unless $x^3 -3xy+y^3 +1= 2$ then the identity on your question is incorrect.
