Answer
$\displaystyle \frac{2y}{\sqrt{x}}$
Work Step by Step
Simplify the radicand by finding as many squared factors as possible:
$4=2^{2}$
$\displaystyle \frac{x}{x^{2}}=\frac{1}{x}$
$\displaystyle \frac{y^{3}}{y}=y^{3-1}=y^{2}$
$\sqrt{\frac{4xy^{3}}{x^{2}y}}=\sqrt{2^{2}\cdot y^{2}\cdot\frac{1}{x}}$
... Radical of a product:
$=\sqrt{2^{2}}\cdot\sqrt{y^{2}}\cdot\sqrt{\frac{1}{x}}$
... for even n, $\sqrt[n]{x^{n}}=|x|$,
$=|2|\cdot|y|\sqrt{\frac{1}{x}}$
...$ y$ is positive, $\quad \displaystyle \sqrt{\frac{1}{x}}=\frac{1}{\sqrt{x}}$
$=\displaystyle \frac{2y}{\sqrt{x}}$
