Answer
$\dfrac{|y| \sqrt[4] {7y}}{|x^3|}$
Work Step by Step
Simplify. $\dfrac{\sqrt[4] {7y^5}}{\sqrt [4] {x^{12}}}$
Now, $\dfrac{\sqrt[4] {7y^5}}{\sqrt [4] {x^{12}}}=\dfrac{\sqrt[4] {7y \cdot y^4}}{\sqrt [4] {(x^3)^{4}}}$
Thus, in the absolute sign,we have
$=\dfrac{|y| \sqrt[4] {7y}}{|x^3|}$