Answer
$f'(x) = -\dfrac{4(3x^2+3)}{3(x^3+3x+9)^{-7/3}}$
Work Step by Step
In order to derivate this function you have to apply the chain rule
Let's make an «u» substitution to make it easier
$u = x^3+3x+9$
$f(u) = u^{-4/3}$
Derivate the function:
$f'(u) = -\dfrac{4}{3}u^{-7/3}u'$
Now let's find u'
$u' = 3x^2+3$
Then undo the substitution, simplify and get the answer:
$f'(x) = -\dfrac{4(3x^2+3)}{3(x^3+3x+9)^{-7/3}}$
