Answer
$y' = \frac{-4x^{3}-3x^{2}+1}{(x^{4}+x^{3}-x-1)^{2}}$
$y' = \frac{-4x^{3}-3x^{2}+1}{((x^2-1)(x^2+x+1))^2}$
Work Step by Step
$y = \frac{1}{(x^{2}-1)(x^{2}+x+1)}$
$y = \frac{1}{x^{4} + x^{3} - x-1}$
$y' = \frac{0-(4x^{3}+3x^{2}-1)}{(x^{4}+x^{3}-x-1)^{2}}$
$y' = \frac{-4x^{3}-3x^{2}+1}{(x^{4}+x^{3}-x-1)^{2}}$
$y' = \frac{-4x^{3}-3x^{2}+1}{((x^2-1)(x^2+x+1))^2}$