Answer
$$\frac{\partial}{\partial x}f(x,y)=\frac{y-x}{(x+y)^3};$$
$$\frac{\partial}{\partial y}f(x,y)=-\frac{2x}{(x+y)^3}.$$
Work Step by Step
1) Regarding $y$ as constant and differentiating with respect to $x$ we get
$$\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\left(\frac{x}{(x+y)^2}\right)=\frac{(x)'_x(x+y)^2-x\left((x+y)^2\right)'_x}{(x+y)^4}=\frac{(x+y)^2-x\cdot 2(x+y)(x+y)'_x}{(x+y)^4}=\frac{(x+y)^2-2x(x+y)}{(x+y)^4}=\frac{(x+y)-2x}{(x+y)^3}=\frac{y-x}{(x+y)^3}$$
2) Regarding $x$ as constant and differentiating with respect to $y$ we get
$$\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\left(\frac{x}{(x+y)^2}\right)=x\frac{\partial}{\partial y}\left(\frac{1}{(x+y)^2}\right)=x\left(-\frac{2}{(x+y)^3}(x+y)'_y\right)=-\frac{2x}{(x+y)^3}.$$
