Reversed Branched logs

By the change of base rule, the first equation can be written as

$\log_{3}(\log_{2}(x)) - \log_{3}(-\log_{2}(y)) = 1 \Longrightarrow \log_{3}\left(-\dfrac{\log_{2}(x)}{\log_{2}(y)}\right) = 1$

$\Longrightarrow -\dfrac{\log_{2}(x)}{\log_{2}(y)} = 3 \Longrightarrow \log_{2}(x) = -3\log_{2}(y) = \log_{2}\left(\dfrac{1}{y^{3}}\right) \Longrightarrow x = \dfrac{1}{y^{3}}.$

Plugging this into the second equation yields $\dfrac{y^{2}}{y^{3}} = 4 \Longrightarrow y = \dfrac{1}{4},$

which in turn gives us that $x = \dfrac{4}{y^{2}} = \dfrac{4}{\frac{1}{16}} = 64.$

Thus $xy = 64*\dfrac{1}{4} = \boxed{16}.$

(Note that the above solutions for $x$ and $y$ do indeed satisfy the first equation.)