Answer
$\sqrt[4] 18y^{3}$
Work Step by Step
The product rule for radicals tells us that $\sqrt[n] a\times\sqrt[n] b=\sqrt[n] ab$ (when $\sqrt[n] a$ and $\sqrt[n] b$ are real numbers and $n$ is a natural number). That is, the product of two nth roots is the nth root of the product.
Therefore, $\sqrt[4] 3y^{2}\times\sqrt[4] 6y=\sqrt[4] (3y^{2}\times 6y)=\sqrt[4] (3\times6\times y^{2}\times y)=\sqrt[4] 18y^{3}$.