$x^{\frac{y}{z}} = y^{\frac{z}{x}} = z^{\frac{x}{y}}$

Raising everything to the power of $xyz$ gives us $x^{xy^{2}} = y^{yz^{2}} = z^{zx^{2}}$ $x=y=z$ is a solution since $x^{x^{3}} = x^{x^{3}} = x^{x^{3}}$ However, $x=y \neq z$ cannot be a solution. To show this, let's first assume that it provides a solution. We then get $x^{x^{3}} = x^{xz^{2}} = z^{zx^{2}}$ Equating the exponents on the first two sides gives $x^{3} = xz^{2}$ , or $z = \pm x$ . $z = - x$ cannot hold because all numbers must be nonzero, while $z = x$ contradicts our condition that $x \neq z$ . Thus, $x=y \neq z$ can never solve this equation.

Nice problem.