Answer
$$\frac{x^{18}y^6}{4}$$
Work Step by Step
\begin{align}
\frac{(2^{-1}x^{-2}y^{-1})^{-2}(2x^{-4}y^3)^{-2}(16x^{-3}y^3)^0}{(2x^{-3}y^{-5})^2} & = \frac{(2^2x^4y^2)(2^{-2}x^8y^{-6})(16^0x^0y^0)}{2^2x^{-6}y^{-10}}\\ &= \frac{2^0x^{12}y^{-4}}{2^2x^{-6}y^{-10}} \\ &= \frac{x^{12}x^6y^{-4}y^{10}}{2^2} \\ &= \frac{x^{18}y^6}{4}
\end{align}