Answer
$\dfrac{2}{27z^{3}}$
Work Step by Step
Using laws of exponents, then,
\begin{array}{l}
\dfrac{2(3yz)^{-3}}{y^{-3}}
\\\\=
\dfrac{2(3^{-3}y^{-3}z^{-3})}{y^{-3}}
\\\\=
\dfrac{2(y^{-3-(-3)})}{3^{3}z^{3}}
\\\\=
\dfrac{2(y^{0})}{3^{3}z^{3}}
\\\\=
\dfrac{2}{27z^{3}}
.\end{array}
