Answer
$4y^{\frac{1}{15}}$
Work Step by Step
Recall the following rule: $a^{n}\cdot a^{m}=a^{n+m}$. Therefore, we obtain:
$(y^{-\frac{1}{3}})(4y^{\frac{2}{5}})$
$=(1\times4)(y^{-\frac{1}{3}+\frac{2}{5}})$
$=(4)(y^{\frac{-1(5)+2(3)}{15}})$
$=4y^{\frac{1}{15}}$
