Find $xyz$ if $x-\frac{1}{x}=y$, and $y-\frac{1}{y}=z$, and $z-\frac{1}{z}=x$ If 
$$x-\frac{1}{x}=y, \qquad y-\frac{1}{y}=z, \qquad z-\frac{1}{z}=x$$ 
find the value of $xyz$.
This is how far I proceeded:
$x+y+z=z-1/z+x-1/x+y-1/y=>1/x+1/y+1/z=0
=>xy+yz+zx=0$
Also from question,
$x^2-1=xy,y^2-1=yz,z^2-1=zx$.
Adding $x^2+y^2+z^2-3=xy+yz+zx=0 =>x^2+y^2+z^2=3$  .
I am stuck here please help.
This image gives some hint, but I am unable to understand it.

Eliminating $x$ and $z$ between the equations we obtain:
$$3y^6-9y^4+6y^2-1=0.$$
By symmetry we know that $x$ and $z$ must also satisfy this equation. Hence $x^2$, $y^2$ and $z^2$ are roots of $$3y^3-9y^2+6y-1=0.$$
Therefore $(xyz)^2=\frac{1}{3}$ and hence $xyz=\pm\frac{1}{\sqrt 3}$.
In addition I think this may be one of the simplest direct solutions:
Adding the equations: $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0=>xy+yz+zx=0.$$ Clearing the denominators of the original equations ($x^2-xy-1=0, \dots$) and adding we obtain $$x^2+y^2+z^2=xy+yz+zx+3=3$$ Hence $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=3=>(x+y+z)=\pm\sqrt{3}$$Rearranging the original equations ($y+\frac{1}{x}=x,\dots$), multiplying them and simplifying: $$(y+\frac{1}{x})(z+\frac{1}{y})(x+\frac{1}{z})=xyz =>(x+y+z)=-\frac{1}{xyz}.$$ Combining we obtain the result $$xyz=\pm\frac{1}{\sqrt{3}}.$$
