See that $$\left(\frac{x^2}y-1\right)\left(\frac{y^2}z-1\right)\left(\frac{z^2}x-1\right)=0$$WLOG $y^2=z$. Then, $1=xyz=xy^3$. Hence; $(x,y,z)=\left(t^3, \frac 1t, \frac 1{t^2}\right)$ for a positive real number $t$. Regardless of the value of $t$, the median of these numbers is $\frac 1t$. Hence, we need to find the minimum value of $\frac{t^3+\frac 1{t^2}}{\frac 1t}=\frac{t^5+1}{t}$. $\left(\frac{t^5+1}{t}\right)’=\frac{4t^5-1}{t^2}$. Hence, this function achieves its minimum value when $t=\sqrt[5]{4}$ and for this value $\frac{t^5+1}{t}=\frac 5{2^{\frac 85}}$, done.