#### Answer

$$\dfrac{dy}{dx}= \frac{2}{3} x^{-4 / 3}$$

#### Work Step by Step

Since $ y=\dfrac{-2}{\sqrt[3]{x}} = -2 x^{-1 / 3}$
Using the rule $\frac{d}{dx}x^n=nx^{n-1},$ then
\begin{align*}
\dfrac{dy}{dx}&= (-2)\frac{-1}{3}x^{-4/3}\\
&=\frac{2}{3} x^{-4 / 3}\end{align*}