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Maximizing $\frac{x+y}{x-y}$ is equivalent to maximizing $$\frac{x+y}{x-y}+1=\frac{2x}{x-y}=\frac{-2x}{y-x}.$$Note that $-2x$ and $y-x$ are always positive, so to maximize this expression, we take $y=2,$ the smallest possible value of $y.$

Then maximizing $\frac{x+2}{x-2}$ is equivalent to maximizing $$\frac{x+2}{x-2}-1=\frac{4}{x-2}=-\frac{4}{2-x}.$$Note that $2-x$ is always positive, so to maximize this expression, we take $x=-5.$ Hence, the maximum value is $\frac{-5+2}{-5-2}=\boxed{\frac{3}{7}}.$