Answer
$$\frac{\partial w}{\partial x}=\frac{7yz}{(x+y)^2}$$
$$\frac{\partial w}{\partial y}=-\frac{7xz}{(x+y)^2}$$
$$\frac{\partial w}{\partial z}=\frac{7x}{x+y}$$
Work Step by Step
We will use quotient rule to find partial derivative with respect to $x$:
$$\frac{\partial w}{\partial x}=\frac{\partial }{\partial x}\left(\frac{7xz}{x+y}\right)=\frac{(x+y)\frac{\partial }{\partial x}(7xz)-7xz\frac{\partial }{\partial x}(x+y)}{(x+y)^2}=\frac{(x+y)\cdot7z-7xz\cdot1}{(x+y)^2}=\frac{7xz+7yz-7xz}{(x+y)^2}=\frac{7yz}{(x+y)^2}$$
Here $y$ and $z$ are treated as a constant.
We will use chain rule to find partial derivative with respect to $y$:
$$\frac{\partial w}{\partial y}=\frac{\partial }{\partial y}\left(\frac{7xz}{x+y}\right)=7xz\frac{\partial }{\partial y}\left(\frac{1}{x+y}\right)=7xz\frac{-1}{(x+y)^2}\frac{\partial }{\partial y}(x+y)=-\frac{7xz}{(x+y)^2}\cdot1=-\frac{7xz}{(x+y)^2}$$
$x$ and $z$ are treated as a constant here.
The partial derivative with respect to $z$ is:
$$\frac{\partial w}{\partial z}=\frac{\partial }{\partial z}\left(\frac{7xz}{x+y}\right)=\frac{7x}{x+y}\frac{\partial }{\partial z}(z)=\frac{7x}{x+y}\cdot1=\frac{7x}{x+y}$$
Because $x$ and $y$ are treated as a constant, $\frac{7x}{x+y}$ is treated as a constant as well.
