Answer
$\frac{y^{\frac{19}{3}}}{z^{\frac{1}{2}}}$
Work Step by Step
$(y^{2}z^{-3})^{\frac{1}{6}}(y^{3})^{2}$
(Raise the factor of $y^{2}z^{-3}$ to the $\frac{1}{6}$ power. Multiply the exponents of a power raised to a power)
$y^{2\times\frac{1}{6}}z^{-3\times\frac{1}{6}}y^{3\times2}$
(Simplify)
$y^{\frac{1}{3}}z^{-\frac{1}{2}}y^{6}$
(Add the exponents of the same base)
$y^{\frac{1}{3}+6}z^{-\frac{1}{2}}$
(Make 6 into a fraction with the same denominator)
$y^{\frac{1}{3}+\frac{18}{3}}z^{-\frac{1}{2}}$
(Simplify)
$y^{\frac{19}{3}}z^{-\frac{1}{2}}$
(Because z has a negative exponent, it must go on the bottom of the fraction)
$\frac{y^{\frac{19}{3}}}{z^{\frac{1}{2}}}$
