Equivalent Equations!

$Given \space , \space x \space + \frac{1}{y} \space =\space y \space + \space \frac{1}{z} \space=\space z \space + \space \frac{1}{x}$

$\rightarrow x \space - \space y \space=\space \frac{1}{z} \space - \space \frac{1}{y} \space=\space \frac{y-z}{yz} \space\space \space \boxed{(1)}$

$\rightarrow y \space - \space z \space=\space \frac{1}{x} \space - \space \frac{1}{z} \space=\space \frac{z-x}{xz} \space\space\space \boxed{(2)}$

$\rightarrow z \space - \space x \space=\space \frac{1}{y} \space - \space \frac{1}{x} \space=\space \frac{x-y}{xy} \space\space\space \boxed{(3)}$

$Multiplying \space (1) \space,\space (2) \space and \space (3) \space, \space we \space get$

$\Rightarrow (x-y)(y-z)(z-x) \space=\space \frac{(x-y)(y-z)(z-x)}{(xyz)^{2}}$

$\boxed{\therefore \space x^{2}y^{2}z^{2} \space=\space 1}$