Answer
$$\frac{\partial z}{\partial x}=\frac{2y}{y^2-x^2}$$
$$\frac{\partial z}{\partial y}=\frac{2x}{x^2-y^2}$$
Work Step by Step
We will use chain rule to find both partial derivatives.
The partial derivative with respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}\left(\ln\left(\frac{x+y}{x-y}\right)\right)=\frac{1}{\frac{x+y}{x-y}}\frac{\partial }{\partial x}\left(\frac{x+y}{x-y}\right)=\frac{x-y}{x+y}\frac{(x-y)\frac{\partial }{\partial x}(x+y)-(x+y)\frac{\partial }{\partial x}(x-y)}{(x-y)^2}=
\frac{x-y}{x+y}\frac{(x-y)\cdot1-(x+y)\cdot1}{(x-y)^2}=\frac{1}{x+y}\frac{x-y-x-y}{x-y}=-2\frac{y}{(x+y)(x-y)}=\frac{2y}{y^2-x^2}$$
Here is $y$ treated as a constant. We used quotient rule to find partial derivative $\frac{\partial }{\partial x}\left(\frac{x+y}{x-y}\right).$
The partial derivative with respect to $y$ is:
$$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}\left(\ln\left(\frac{x+y}{x-y}\right)\right)=\frac{1}{\frac{x+y}{x-y}}\frac{\partial }{\partial y}\left(\frac{x+y}{x-y}\right)=\frac{x-y}{x+y}\frac{(x-y)\frac{\partial }{\partial y}(x+y)-(x+y)\frac{\partial }{\partial y}(x-y)}{(x-y)^2}
=\frac{x-y}{x+y}\frac{(x-y)\cdot1-(x+y)\cdot(-1)}{(x-y)^2}=\frac{1}{x+y}\frac{x-y+x+y}{x-y}=\frac{2x}{x^2-y^2}$$
because now $x$ is treated as a constant. We also used quotient rule to find partial derivative $\frac{\partial }{\partial y}\left(\frac{x+y}{x-y}\right).$
