Answer
$f'(x)=\dfrac{-1-x^2}{(x^2-1)^2}.$
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=x^2+x-1; u'(x)=2x+1$
$v(x)=x^2-1; v'(x)=2x$
$f'(x)=\dfrac{(2x+1)(x^2-1)-(2x)(x^2+x-1)}{(x^2-1)^2}$
$=\dfrac{-1-x^2}{(x^2-1)^2}.$