jmc

We have $|-2y|\cdot \left|-\frac{y}{2}\right|=\left|\frac{2y^2}{2}\right|=|y^2|$. Because $y^2>0$, we have $|y^2|=y^2$, so our original equation becomes $y^2=36$. Therefore, we have $y=6$ or $y=-6$, and the product of these is $\boxed{-36}$.