Answer
$\dfrac{1}{x^{7/4}}$
Work Step by Step
Using laws of exponents, then,
\begin{array}{l}
\dfrac{(x^3y^2)^{1/4}}{(x^{-5}y^{-1})^{-1/2}}
\\\\=
\dfrac{x^{3\cdot\frac{1}{4}}y^{2\cdot\frac{1}{4}}}{x^{-5\cdot-\frac{1}{2}}y^{-1\cdot-\frac{1}{2}}}
\\\\=
\dfrac{x^{\frac{3}{4}}y^{\frac{2}{4}}}{x^{\frac{5}{2}}y^{\frac{1}{2}}}
\\\\=
x^{\frac{3}{4}-\frac{5}{2}}y^{\frac{2}{4}-\frac{1}{2}}
\\\\=
x^{\frac{3}{4}-\frac{10}{4}}y^{\frac{2}{4}-\frac{2}{4}}
\\\\=
x^{-\frac{7}{4}}y^{0}
\\\\=
\dfrac{1}{x^{7/4}}
.\end{array}
