I have to prove that if $\displaystyle |x| <> 1, y <> 0, y<>-x$ then identity is valid:

$\displaystyle \frac{x^2-1}{y^2+xy}\cdot(\frac{1}{1-\frac{1}{y}}-1)\cdot\frac{1-xy^3-y^4+y}{1-x^2}=\frac{y^2+y+1}{y}$

I have try to prove it.This what I have done:

$\displaystyle \frac{x^2-1}{y^2+xy}\cdot(\frac{1}{y-1})\cdot\frac{1-xy^3-y^4+y}{1-x^2}=\frac{y^2+y+1}{y}$

$\displaystyle \frac{(1-x^2)(-1+xy^3+y^4-y)}{(y^2+xy)(y-1)(1-x^2)}=\frac{y^2+y+1}{y}$

$\displaystyle \frac{(-1+xy^3+y^4-y)}{(y^2+xy)(y-1)}=\frac{y^2+y+1}{y}$

$\displaystyle \frac{-1+xy^3+y^4-y}{y^3-y^2+xy^2-xy}=\frac{y^2+y+1}{y}$

$\displaystyle \frac{y(-1+xy^3+y^4-y)}{y(y^2-y+xy-x)}=y^2+y+1$

$\displaystyle \frac{-1+xy^3+y^4-y}{y^2-y+xy-x}=y^2+y+1$

$\displaystyle -1+xy^3+y^4-y=(y^2-y+xy-x)(y^2+y+1)$

$\displaystyle -1+xy^3+y^4-y=-x+xy^3+y^4-y$

$\displaystyle x=1$

So identity is not valid.

Am I right?