9. $$\frac{4}{x-y}$$ - $$\frac{3}{x+y}$$ - $$\frac{8y}{x^2-y^2}$$ = $$\frac{4(x+y)-3(x-y)-8y}{(x-y)(x+y)}$$ = $$\frac{4x+4y-3x+3y-8y}{x^2-y^2}$$ = $$\frac{x-y}{x^2-y^2}$$ = $$\frac{x-y}{(x-y)(x+y)}$$ = $$\frac{1}{x+y}$$
Ответ: $$\frac{1}{x+y}$$