Answer
$3x^3y^4 \sqrt[3] {2x^2y}$
Work Step by Step
Apply product rule: $\sqrt[n] p \sqrt[n] q=\sqrt[n]{pq}$
Here, $n$ refers as index.
$\sqrt[3] {6x^7y} \sqrt[3] {9x^4y^{12}}=\sqrt[3] {(6x^7y) (9x^4y^{12})}= \sqrt[3]{54x^{11}y^{13}}$
The radical $\sqrt[3]{54x^{11}y^{13}}$ can be further simplified by using product rule again.
Such as:
$\sqrt[3]{54x^{11}y^{13}}=\sqrt[3] {(3x^3y^4)^3 (2x^2y)}=\sqrt[3] {(3x^3y^4)^3} \sqrt[3]{ (2x^2y)}=3x^3y^4 \sqrt[3] {2x^2y}$
