Answer
(a) $\frac{2y^{4/3}}{x^{2}}$
(b) $-\frac{y^{3/4_{Z}2}}{2}$
Work Step by Step
(a) $(\frac{x^{8}y^{-4}}{16y^{4/3}})^{-1/4}=16^{-1*-1/4}x^{8*-1/4}*y^{(-4)(-1/4)-(4/3)*(-1/4)}=\frac{2y^{4/3}}{x^{2}}$
(b) $(\frac{-8y^{3/4}}{y^{3}z^{6}})^{-1/3}=(-8)^{-1/3}y^{3/4(-1/3)-3(-1/3)}z^{-6*-1/3}=-\frac{y^{3/4_{Z}2}}{2}$
