Answer
$27y^{\frac{2}{3}}$
Work Step by Step
$\displaystyle \frac{(3y^{1/4})^{3}}{y^{1/12}}=\qquad$... apply $(ab)^{n}=a^{n}b^{n}$
$=\displaystyle \frac{(3^{3})(y^{1/4})^{3}}{y^{1/12}}$... apply $(a^{m})^{n}=a^{mn}$
$=\displaystyle \frac{27y^{3/4}}{y^{1/12}}\qquad$... apply $\displaystyle \frac{a^{m}}{a^{n}}=a^{m-n}$
...$ \displaystyle \frac{3}{4}-\frac{1}{12}=\frac{9-1}{12}=\frac{8}{12}=\frac{2}{3}$
$=27y^{\frac{2}{3}}$
