Is that the correct answer to the problem?
I got in radicals: 6 and 11, $\displaystyle x^6, y{^14}, y$
No, it isn't right. You forgot to take the square roots of $\displaystyle x^6$ and $\displaystyle y^{14}$.
$\displaystyle \sqrt{396x^6y^{15}}$
$\displaystyle = \sqrt{36x^6y^{14} \cdot 11y}$
$\displaystyle = \sqrt{36} \sqrt{x^6} \sqrt{y^14} \sqrt{11y}$
$\displaystyle = 6x^3y^7 \sqrt{11y}$
-Dan