Definitely not $x=y=z$

$y=\frac{2z^2}{x+z}, (2x \leq z)$

Let the term be F, we have

$F=\frac{x}{x-\frac{2z^2}{x+z}}+\frac{\frac{2z^2}{x+z}}{\frac{2z^2}{x+z}-z}+\frac{z}{z-x}$ $=\frac{x(x+z)}{(x-z)(x+2z)}-\frac{2z}{x-z} -\frac{z}{z-x}$ $=\frac{x^2-2xz-6z^2}{x^2+xz-2z^2}$

By the $Homogenity$ of $F$ , we can set $t=\frac{x}{z}$

So,

$F(t)=\frac{t^2-2t-6}{t^2+t-2}$

We need to find stationary value of F,

So, we need to determine the range of $t$ ,

Because $x,z >0$ and $2x \leq z$ , we have $0 \leq t \leq \frac{1}{2}$

Checking whether which stationary point between $0 \leq t \leq \frac{1}{2}$

$\frac{dF}{dt}=0$ $\frac{3t^2+16t+2}{(t+2)^2(t-1)^2}=0$ $t=\frac{-8+\sqrt{58}}{3},\frac{-8-\sqrt{58}}{3}$ $t=-0.128,-5.205$

It seems these two values aren't between $0\leq t \leq \frac{1}{2}$

So $F$ is monotonous between $0 \leq t \leq \frac{1}{2}$

Calculating $F(0)$ and $F(\frac{1}{2})$ , we get

$F(0)=3$ $F(\frac{1}{2})=5.4$

Which leads $F_{Max}=\boxed{5.4000}$ and give us $F$ is monotonously increasing between $0\leq t \leq\ \frac{1}{2}$

Feel free to tell me if I have any wrong :)

Quick and dirty ES6:

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// Our function
var mystery = (x,y,z) => (x/(x-y) + y/(y-z) + z/(z-x));

// y solved for x,z
var y_xz = (x,z) => (2*z*z/(x+z));

var stepsize = 0.1;
var maxV = -1;
for(let z=0; z<=100; z+=stepsize){
    for(let x=0; x <= z/2; x+=stepsize){
        let y=y_xz(x,z);
        let cache = mystery(x,y,z);

        if(cache > maxV){
            maxV = cache;
            console.log(x,y,z,maxV);
        }
    }
}
console.log(maxV);

Output:

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0 0.20000000000000004 0.1 3
0.1 0.26666666666666666 0.2 5.4
5.4

After looking at the last max, searching within $z\leq100$ gave me reasonable confidence that 5.4 the right answer.