Wrong solution - but why ? 
Find all solutions to the ODE $$y'=\begin{pmatrix}0 & 1 \\ \frac{-2}{1-x^2} & \frac{2x}{1-x^2}\end{pmatrix}y$$
  What I did :
  Guessing $y_1=\begin{pmatrix}x \\1\end{pmatrix}$ and reduce the order:
  complete $y_1$ to an invertible matrix such that $$H^{-1}=\begin{pmatrix}0 & 1 \\ 1 & -x\end{pmatrix}$$
  Calculate $$B=\begin{pmatrix}0 & 1 \\ 1 & -x\end{pmatrix}  \begin{pmatrix}0 & 1 \\ \frac{-2}{1-x^2} & \frac{2x}{1-x^2}\end{pmatrix}\begin{pmatrix}1 \\0\end{pmatrix}-\begin{pmatrix}0 \\0\end{pmatrix}$$
  So $$B=\frac{2}{x^2-1}\begin{pmatrix}1 \\-x\end{pmatrix}$$ and 
$B_1=\frac{2}{x^2-1}$ and $B_2=\frac{-2x}{x^2-1}$
  solve $z'=B_2z$
$\Rightarrow C_2=x^2-1$ and
  $C_1=\int B_1 C_2dx = \int 2dx=2x$
Calculate $HC$
  $$HC=\begin{pmatrix}x & 1 \\ 1 & 0\end{pmatrix} \begin{pmatrix}2x  \\ x^2-1\end{pmatrix}=\begin{pmatrix}3x^2-1  \\ 2x\end{pmatrix}$$

This should be an soultion but it doesn't work and I don't know why :(

In the wording of the question some symbols are undefined such as $B$, $C_1$ , $C_2$ , $z\quad$ This is confusing and makes difficult to answer with any certainty. 
Nevertheless, the end of calculus might be :
$$z'=-\frac{2x}{x^2-1}z \quad\to\quad z=\frac{c_2}{x^2-1}$$
$$\int \frac{1}{(x^2-1)^2}dx=\frac{1}{2}\ln\left|\frac{x+1}{x-1}\right|-\frac{x}{x^2-1}$$
$$y=c_1\left(\begin{matrix}x\\1 \end{matrix}\right)+c_2\left(\begin{matrix}\frac{1}{2}x\ln\left|\frac{1+x}{1-x}\right|-1\\ \frac{1}{2}\ln\left|\frac{1+x}{x-1}\right|-\frac{x}{x^2-1} \end{matrix}\right)$$
