Answer
Both -1 and 1 are extraneous solutions. There is no solution.
Work Step by Step
$$\frac{x}{x-1}-\frac{1}{x+1}=\frac{2}{x^2-1}$$
$$((x+1)(x-1))(\frac{x}{x-1}-\frac{1}{x+1})=((x+1)(x-1))(\frac{2}{x^2-1})$$
$$x(x+1)-(x-1)=2$$
$$x^2+x-x+1=2$$
$$x^2-1=0$$
$$(x+1)(x-1)=0$$
$$x=-1,1$$
Both -1 and 1 are extraneous solutions. There is no solution.