#### Answer

Both 1 and -1 are extraneous solutions. There is no solution.

#### Work Step by Step

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