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The difference between consecutive terms is $(x-y)-(x+y)=-2y.$ Therefore we can also express the third and fourth terms as $x-3y$ and $x-5y.$ Then we can set them equal to $xy$ and $\frac{x}{y}$ because they are the same thing. $$\begin{array}{rl}xy& =x-3y\\ xy-x& =-3y\\ x(y-1)& =-3y\\ x& =\frac{-3y}{y-1}\end{array}$$ Substitute into our other equation. $$\frac{x}{y}=x-5y$$ $$\frac{-3}{y-1}=\frac{-3y}{y-1}-5y$$ $$-3=-3y-5y(y-1)$$ $$0=5y^2-2y-3$$ $$0=(5y+3)(y-1)$$ $$y=-\frac{3}{5},1$$ But $y$ cannot be $1$ because then the first term would be $x+1$ and the second term $x-1$ while the last two terms would be equal to $x.$ Therefore $y=-\frac{3}{5}.$ Substituting the value for $y$ into any of the equations, we get $x=-\frac{9}{8}.$ Finally, $$\frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\frac{123}{40}}$$