Answer
$\dfrac{15}{8y^{5}z^{3}}$
Work Step by Step
Using the laws of exponents, the given expression, $
\dfrac{(-5y^3z^4)^2(2yz^5)^{-2}}{10(y^4z)^{3}(3y^3z^2)^{-1}}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{(-5)^2y^{3(2)}z^{4(2)}(2)^{-2}y^{-2}z^{5(-2)}}{10y^{4(3)}z^{3}(3)^{-1}y^{3(-1)}z^{2(-1)}}
\\\\=
\dfrac{(-5)^2(3)^{1}y^{6}z^{8}y^{-2}z^{-10}}{10(2)^{2}y^{12}z^{3}y^{-3}z^{-2}}
\\\\=
\dfrac{25(3)y^{6+(-2)}z^{8+(-10)}}{10(4)y^{12+(-3)}z^{3+(-2)}}
\\\\=
\dfrac{75y^{4}z^{-2}}{40y^{9}z^{1}}
\\\\=
\dfrac{\cancel{5}(15)y^{4-9}z^{-2-1}}{\cancel{5}(8)}
\\\\=
\dfrac{15y^{-5}z^{-3}}{8}
\\\\=
\dfrac{15}{8y^{5}z^{3}}
.\end{array}