Answer
$ -\dfrac{2\sqrt[3] 3}{3}$
Work Step by Step
Rationalize the denominator by multiplying $\sqrt[3]3$ to both the numerator and the denominator:
$\dfrac{-2}{\sqrt[3] 9}=\dfrac{-2}{\sqrt[3] 9}\cdot \dfrac{\sqrt[3] 3}{\sqrt[3] 3}$
Use the rule $\sqrt[n]{a} \cdot \sqrt[n]{b}=\sqrt[n]{a\cdot b}$ then simplify:
$= \dfrac{-2\sqrt[3] 3}{\sqrt[3] {9\cdot 3}}$
$= \dfrac{-2\sqrt[3] 3}{\sqrt[3] {27}}$
$= \dfrac{-2\sqrt[3] 3}{\sqrt[3] {3^3}}$
$= \dfrac{-2\sqrt[3] 3}{3}$
Hence, the correct answer is $ -\frac{2\sqrt[3] 3}{3}$.
