jmc

First, note that $$\begin{array}{rl}f(x,y,z)& =\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\\ & >\frac{x}{x+y+z}+\frac{y}{y+z+x}+\frac{z}{z+x+y}\\ & =\frac{x+y+z}{x+y+z}=1.\end{array}$$Let $ϵ$ be a small positive number. Then $$\begin{array}{rl}f({ϵ}^2,ϵ,1)& =\frac{{ϵ}^2}{{ϵ}^2+ϵ}+\frac{ϵ}{ϵ+1}+\frac{1}{1+{ϵ}^2}\\ & =\frac{ϵ}{ϵ+1}+\frac{ϵ}{ϵ+1}+\frac{1}{1+{ϵ}^2}.\end{array}$$As $ϵ$ approaches 0, $f({ϵ}^2,ϵ,1)$ approaches 1. This means we can make $f(x,y,z)$ arbitrarily close to 1, without actually reaching 1.

Now, note that $$f(x,y,z)+f(x,z,y)=\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}+\frac{x}{x+z}+\frac{z}{z+y}+\frac{y}{x+y}=3.$$Therefore, $f(x,y,z)<2,$ and we can make $f(x,y,z)$ arbitrarily close to 2.

Hence, the set of all possible values of $f(x,y,z)$ is $\boxed{(1,2)}.$