The first equation $\frac{x-y}{x+y}=9$ gives $x-y=9x+9y$ and so $-8x=10y$ or $-4x=5y$. The second equation $\frac{xy}{x+y}=-60$ gives $xy=-60x-60y$. Multiplying this equation by 5 gives $5xy=-300x-300y$ or $x(5y)=-300x-60(5y)$. Since $5y=-4x$, then $x(-4x)=-300x-60(-4x)$ or $-4x^{2}=-60x$. Rearranging, we obtain $4x^{2}-60x=0$ or $4x(x-15)=0$. Therefore, $x=0$ or $x=15$. Since $y=-\frac{4}{5}x$, then $y=0$ or $y=-12$. From the first equation, it cannot be the case that $x=y=0$. We can check that the pair $(x, y)=(15,-12)$ satisfies both equations. Therefore, $(x+y)+(x-y)+xy=3+27+(-180)=-150$.