Answer
$\dfrac{b^{5/8}}{a}$
Work Step by Step
Using laws of exponents, then,
\begin{array}{l}
\dfrac{(a^{-2}b^{3})^{1/8}}{(a^{-3}b)^{-1/4}}
\\\\=
\dfrac{a^{-2\cdot\frac{1}{8}}b^{3\cdot\frac{1}{8}}}{a^{-3\cdot-\frac{1}{4}}b^{-\frac{1}{4}}}
\\\\=
\dfrac{a^{\frac{-2}{8}}b^{\frac{3}{8}}}{a^{\frac{3}{4}}b^{-\frac{1}{4}}}
\\\\=
a^{\frac{-2}{8}-\frac{3}{4}}b^{\frac{3}{8}-\frac{-1}{4}}
\\\\=
a^{\frac{-2}{8}-\frac{6}{8}}b^{\frac{3}{8}+\frac{2}{8}}
\\\\=
a^{\frac{-8}{8}}b^{\frac{5}{8}}
\\\\=
\dfrac{b^{5/8}}{a}
.\end{array}
