jmc

Let $\frac{x}{y}=t$. Then $x=ty$, so we can write $$\frac{x-y}{x+y}=\frac{ty-y}{ty+y}=\frac{t-1}{t+1}.$$Thus, we have $$2<\frac{t-1}{t+1}<5,$$which we can rewrite as follows: $$\begin{array}{rl}2<1& -\frac{2}{t+1}<5\\ 1<& -\frac{2}{t+1}<4\\ -\frac{1}{2}>& \frac{1}{t+1}>-2.\end{array}$$The only number of the form $\frac{1}{t+1}$ (where $t$ is an integer) which lies in the interval $\left(-2,-\frac{1}{2}\right)$ is $-1=\frac{1}{-1}$, so we must have $t+1=-1$, and $t=-2$. This is achievable when $x=-2$ and $y=1$, so the answer is $\boxed{-2}$.