Answer
$$\dfrac{|3x|}{y^2}$$
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[4]{162x^4}}{\sqrt[4]{2y^8}}&=\sqrt[4]{\dfrac{162x^4}{2y^8}}\\\\
&=\sqrt[4]{\dfrac{\cancel{162}^{81}x^{4}}{\cancel{2}y^8}}\\\\
&=\sqrt[4]{\dfrac{81x^4}{y^8}}\\\\
&=\sqrt[4]{\dfrac{(3x)(3x)(3x)(3x)}{(y^2)(y^2)(y^2)(y^2)}}\\\\
&=\sqrt[4]{\dfrac{\left(3x\right)^4}{\left(y^2\right)^4}}\\\\
&=\sqrt[4]{\left(\dfrac{3x}{y^2}\right)^4}
\end{align*}
Use the rule $\sqrt[n]{a^n} = |a|$ where $a$ is even to obtain:
$$=\left|\dfrac{3x}{y^2}\right|$$
However, since $y^2$ is never negative, then $|y^2|=y^2$.
Hence,
$$\left|\dfrac{3x}{y^2}\right|=\dfrac{|3x|}{y^2}$$
