Answer
$f^{\prime}(x)=3(x^{3}-x)^{2}\cdot(3x^{2}-1)$
Work Step by Step
Let $u(x)=x^{3},\qquad v(x)=x^{3}-x$
$\displaystyle \frac{du}{dx}=3x^{2}, \quad \frac{dv}{dx}=3x^{2}-1$
Then,$\quad y=f(x)=u(v(x))$ and $f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dv}=3v^{2}=3(x^{3}-x)^{2}$
$f^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=3(x^{3}-x)^{2}\cdot(3x^{2}-1)$
$f^{\prime}(x)=3(x^{3}-x)^{2}\cdot(3x^{2}-1)$
