Find $xyz$ if $x-\frac{1}{x}=y$, and $y-\frac{1}{y}=z$, and $z-\frac{1}{z}=x$

Given: $x-\frac1x=y,y-\frac1y=z,z-\frac1z=x$, you found correctly: $$x-\frac1x+y-\frac1y+z-\frac1z=y+z+x \Rightarrow \frac1x+\frac1y+\frac1z=0 \Rightarrow xy+yz+zx=0 \ \ (1)\\ x^2-1+y^2-1+z^2-1=xy+yz+zx \Rightarrow x^2+y^2+z^2=3 \ \ (2)$$ Square each and add them all: $$x^2+\frac1{x^2}-2+y^2+\frac1{y^2}-2+z^2+\frac1{z^2}-2=y^2+z^2+x^2 \Rightarrow \frac1{x^2}+\frac1{y^2}+\frac1{z^2}=6 \Rightarrow \\ x^2y^2+y^2z^2+z^2x^2=6x^2y^2z^2 \ \ (3)$$ Square $(1)$: $$\underbrace{x^2y^2+y^2z^2+z^2x^2}_{6x^2y^2z^2}+2xyz(x+y+z)=0 \Rightarrow x+y+z=-3xyz \ \ (4)$$ Square $(4)$: $$\underbrace{x^2+y^2+z^2}_{3}+2(\underbrace{xy+yz+zx}_{0})=9x^2y^2z^2 \Rightarrow 3=9x^2y^2z^2 \Rightarrow xyz=\pm \frac1{\sqrt{3}}.$$