Answer
$\color{blue}{\dfrac{2x\sqrt[4]{2xy^3}}{y^2}}$
Work Step by Step
Rationalize the denominator by multiplying $y^3$ to both the numerator and denominator of the radicand to obtain:
$=\sqrt[4]{\dfrac{32x^5(y^3)}{y^5(y^3)}}
\\=\sqrt[4]{\dfrac{32x^5y^3}{y^8}}
\\=\sqrt[4]{\dfrac{32x^5y^3}{(y^2)^4}}$
Bring out the fourth root of the denominator to obtain:
$\\=\dfrac{\sqrt[4]{32x^5y^3}}{y^2}
\\=\dfrac{\sqrt[4]{2^5x^5y^3}}{y^2}$
Factor the radicand such that at least one factor is a perfect fourth power to obtain:
$\\=\dfrac{\sqrt[4]{2^4x^4(2xy^3)}}{y^2}$
Bring out the fourth root of the perfect fourth power factor/s of the numerator to obtain:
$\\=\color{blue}{\dfrac{2x\sqrt[4]{2xy^3}}{y^2}}$
