Answer
$f'(x)=-\dfrac{2x}{3\sqrt[3] {(x^{2}-1)^{4}}}$
Work Step by Step
$f(x)=\dfrac{1}{\sqrt[3] {x^{2}-1}}$
Let's do some algebra to change the appearance of the function:
$f(x)=\dfrac{1}{\sqrt[3] {x^{2}-1}}=\dfrac{1}{(x^{2}-1)^{1/3}}=(x^{2}-1)^{-1/3}$
Now, differentiate using the chain rule:
$f'(x)=-\dfrac{1}{3}(x^{2}-1)^{-4/3}(x^{2}-1)'=...$
$...=-\dfrac{1}{3}(x^{2}-1)^{-4/3}(2x)=-\dfrac{2}{3}x(x^{2}-1)^{-4/3}=...$
$...=-\dfrac{2x}{3(x^{2}-1)^{4/3}}=-\dfrac{2x}{3\sqrt[3] {(x^{2}-1)^{4}}}$
