Answer
$\dfrac{y\sqrt[3]{150x}}{10x^2}$
Work Step by Step
Recall:
If $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers, then $\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\dfrac{a}{b}},$ where $a\ne0$.
Use the rule above to obtain:
\begin{align*}
\require{cancel}
\dfrac{\sqrt[3]{6x^2y^4}}{2\sqrt[3]{5x^7y}}&=\dfrac{1}{2} \cdot \dfrac{\sqrt[3]{6x^2y^4}}{\sqrt[3]{5x^7y}}\\\\
&=\dfrac{1}{2}\cdot \sqrt[3]{\dfrac{6x^2y^4}{5x^7y}}\\\\
&=\dfrac{1}{2}\cdot \sqrt[3]{\dfrac{6x^{\cancel{2}}y^{\cancel{4}^3}}{5x^{\cancel{7}^6}\cancel{y}}}\\\\
&=\dfrac{1}{2}\cdot \sqrt[3]{\dfrac{6xy^{3}}{5x^{^6}}}\\\\
&=\dfrac{1}{2}\cdot \sqrt[3]{\dfrac{6x}{5}\cdot \dfrac{y^3}{x^6}}\\\\
&=\dfrac{1}{2}\cdot \sqrt[3]{\dfrac{6x}{5}\cdot \left(\dfrac{y}{x^2}\right)^3}\\\\
\end{align*}
Use the rule $\sqrt[n]{a^n} = a$ where $n$ is even to obtain:
\begin{align*}
&=\dfrac{1}{2}\cdot \dfrac{y}{x^2}\cdot \sqrt[3]{\frac{6x}{5}}\\\\
&=\dfrac{y}{2x^2} \cdot \sqrt[3]{\frac{6x}{5}}
\end{align*}
Rationalize the denominator by multiplying $25$ to both the numerator and denominator of the radicand (expressions inside the radical sign) to obtain:
\begin{align*}
&=\dfrac{y}{2x^2} \cdot \sqrt[3]{\frac{6x}{5} \cdot \frac{25}{25}}\\\\
&=\dfrac{y}{2x^2} \cdot \sqrt[3]{\frac{150x}{125}}\\\\
&=\dfrac{y}{2x^2} \cdot \sqrt[3]{\frac{150x}{5^3}}\\\\
&=\dfrac{y}{2x^2} \cdot \dfrac{1}{5} \sqrt[3]{150x}\\\\
&=\dfrac{y\sqrt[3]{150x}}{10x^2}
\end{align*}
