#### Answer

$f'(t)=\dfrac{-2t-3}{3\sqrt[3] {t^2}(t-3)^2}$

#### Work Step by Step

$f(t)=\dfrac{\sqrt[3] t}{t-3}$
Convert radical to exponent form:
$f(t)=\dfrac{t^{1/3}}{t-3}$
Use the quotient rule:
$f'(t)=\dfrac{(t-3)(\frac{1}{3}t^{-2/3})-(t^{1/3})(1)}{(t-3)^2}$
Distribute the numerator:
$f'(t)=\dfrac{\frac{1}{3}t^{1/3}-t^{-2/3}-t^{1/3}}{(t-3)^2}$
Combine like terms:
$f'(t)=\dfrac{-\frac{2}{3}t^{1/3}-t^{-2/3}}{(t-3)^2}$
Use the definition of a negative exponent and convert back to radical form:
$f'(t)=\dfrac{-\frac{2\sqrt[3] t}{3}-\frac{1}{\sqrt[3] {t^2}}}{(t-3)^2}$
Common denominator:
$f'(t)=\dfrac{\frac{-2t-3}{3\sqrt[3] {t^2}}}{(t-3)^2}$
Make into one fraction:
$f'(t)=\dfrac{-2t-3}{3\sqrt[3] {t^2}(t-3)^2}$