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Minimizing $\frac{x+y}{x-y}$ is equivalent to minimizing $$\frac{x+y}{x-y}+1=\frac{2x}{x-y}=\frac{-2x}{y-x}.$$Note that $-2x$ and $y-x$ are always positive, so to minimize this expression, we take $y=4,$ the largest possible value of $y.$

Then minimizing $\frac{x+4}{x-4}$ is equivalent to minimizing $$\frac{x+4}{x-4}-1=\frac{8}{x-4}=-\frac{8}{4-x}.$$Note that $4-x$ is always positive, so to minimize this expression, we take $x=-3.$ Hence, the minimum value is $\frac{-3+4}{-3-4}=\boxed{-\frac{1}{7}}.$