#### Answer

$\frac{2\sqrt[3] {3x^{2}}}{3x}$

#### Work Step by Step

First, we simplify the expression:
$\sqrt[3] \frac{8}{9x}=\frac{\sqrt[3] {2^{3}}}{\sqrt[3] {9x}}=\frac{2}{\sqrt[3] {9x}}$.
Now, we multiply the expression by $\sqrt[3] {3x^{2}}$ to rationalize the denominator:
=$\frac{2}{\sqrt[3] {9x}}\times\frac{\sqrt[3] {3x^{2}}}{\sqrt[3] {3x^{2}}}$
=$\frac{2\times\sqrt[3] {3x^{2}}}{\sqrt[3] {9x}\times\sqrt[3] {3x^{2}}}$
=$\frac{2\sqrt[3] {3x^{2}}}{\sqrt[3] {9x\times3x^{2}}}$
=$\frac{2\sqrt[3] {3x^{2}}}{\sqrt[3] {27x^{3}}}$
=$\frac{2\sqrt[3] {3x^{2}}}{3x}$