Answer
$\frac{\sqrt[3] (2x)}{3y^{4}\sqrt[3] 3}$
Work Step by Step
The quotient rule holds that $\sqrt[n] (\frac{a}{b})=\frac{\sqrt[n] a}{\sqrt[n] b}$ (where $\sqrt[n] a$ and $\sqrt[n] b$ are real numbers and $\sqrt[n] b$ is nonzero).
Therefore, $\sqrt[3] (\frac{2x}{81y^{12}})=\frac{\sqrt[3] (2x)}{\sqrt[3] (81y^{12})}=\frac{\sqrt[3] (2x)}{\sqrt[3] ((27\times3)y^{12})}=\frac{\sqrt[3] (2x)}{\sqrt[3] (27y^{12})\times\sqrt[3] 3}=\frac{\sqrt[3] (2x)}{3y^{4}\sqrt[3] 3}$
We know that $\sqrt[3] (27y^{12})=3y^{4}$, because $(3y^{4})^{3}=(3\times3\times3)\times y^{4+4+4}=27y^{12}$.
