Answer
$$\frac{\partial f(3,1,-1)}{\partial x}=0$$
$$\frac{\partial f(3,1,-1)}{\partial y}=\frac{2}{3}$$
$$\frac{\partial f(3,1,-1)}{\partial x}=-\frac{1}{3}$$
Work Step by Step
The partial derivative with respect to $x$ is:
$$\frac{\partial f(x,y,z)}{\partial x}=\frac{\partial }{\partial x}\left(\frac{xy}{x+y+z}\right)=\frac{\frac{\partial}{\partial x}(xy)(x+y+z)-xy\frac{\partial}{\partial x}(x+y+z)}{(x+y+z+)^2}=
\frac{y(x+y+z)-xy\cdot1}{(x+y+z)^2}=\frac{y^2+yz}{(x+y+z)^2}$$
The partial derivative with respect to $y$ is:
$$\frac{\partial f(x,y,z)}{\partial y}=\frac{\partial }{\partial y}\left(\frac{xy}{x+y+z}\right)=\frac{\frac{\partial}{\partial y}(xy)(x+y+z)-xy\frac{\partial}{\partial y}(x+y+z)}{(x+y+z+)^2}=
\frac{x(x+y+z)-xy\cdot1}{(x+y+z)^2}=\frac{x^2+xz}{(x+y+z)^2}$$
The partial derivative with respect to $z$ is:
$$\frac{\partial f(x,y,z)}{\partial z}=\frac{\partial }{\partial z}\left(\frac{xy}{x+y+z}\right)=xy\frac{\partial}{\partial z}((x+y+z)^{-1})=xy\cdot(-1)(x+y+z)^{-2}=-\frac{xy}{(x+y+z)^2}$$
Now we will evaluate these partial derivatives at the given point $(x,y,z)=(3,1,-1)$:
$$\frac{\partial f(3,1,-1)}{\partial x}=\left.\frac{y^2+yz}{(x+y+z)^2}\right|_{(x,y,z)=(3,1,-1)}=
\frac{1^2+1\cdot(-1)}{(3+1-1)^2}=0$$
$$\frac{\partial f(3,1,-1)}{\partial y}=\left.\frac{x^2+xz}{(x+y+z)^2}\right|_{(x,y,z)=(3,1,-1)}=\frac{3^2+3\cdot(-1)}{(3+1-1)^2}=\frac{6}{9}=\frac{2}{3}$$
$$\frac{\partial f(3,1,-1)}{\partial z}=\left.-\frac{xy}{(x+y+z)^2}\right|_{(x,y,z)=(3,1,-1)}=-\frac{3\cdot1}{(3+1-1)^2}=-\frac{1}{3}$$
