Konigsberg wrote: Solve the system of equations: $x^2=\frac{1}{y}+\frac{1}{x}$, $y^2=\frac{1}{z}+\frac{1}{x}$, $z^2=\frac{1}{x}+\frac{1}{y}$. in the real numbers. Subtracting first from third, we get $x^2=z^2$ and so $x=z$ (since $x=-z$ would imply in second equation $y=0$, and so first and third undefined). Then second equation gives $x=\frac 2{y^2}$ and first equation becomes $\frac 4{y^4}=\frac 1y+\frac {y^2}2$ And so $y^6+2y^3-8=0$ and $y^3\in\{-4,2\}$ Hence the result : $\boxed{(x,y,z)\in\left\{(\sqrt[3]2,\sqrt[3]2,\sqrt[3]2),\left(\frac 1{\sqrt[3]2},-\sqrt[3]4,\frac 1{\sqrt[3]2}\right)\right\}}$

Oops! I've made a horrible typo! The first equation is supposed to read $x^2=\frac{1}{y}+\frac{1}{z}$. Original Post Corrected.

Clearly $xyz\ne 0.$ Subtracting the second equation from the first we get $(x-y)\left(x+y-\frac{1}{xy}\right)=0.\ (1)$ If $x=y,$ then the system is equivalent to $x^2-\frac{1}{x}=\frac{1}{z}\ (2)$ and $x=\frac{2}{z^2}.\ (3)$ Substituting $(3)$ into $(2)$ we get the quadratic $z^6+2z^3-8=0,$ which gives $z^3=2$ or $z^3=-4.$ Therefore, the solutions are $(x,y,z)=(2^{1/3},2^{1/3},2^{1/3})$ and $(x,y,z)=(2^{-1/3},2^{-1/3},-2^{2/3})$ (and cyclic permutations). Now assume that no two of $x,y,z$ are equal. Therefore, from $(1)$ we get $x+y=\frac{1}{xy}.$ Similarly, by subtracting the third from the first equation, we get $x+z=\frac{1}{xz}.$ Subtracting these two equations we get $xyz=-1.\ (4)$ Substituting into any of the original equations we get $x+y+z=0.\ (5)$ Therefore, the original system is now equivalent to equations $(4),(5).$ Thus, for a given $x\ne 0$ such that $x^4+4x\ge 0,$ the solution is $(x,y,z)=\left(x,\frac{-x^2\pm \sqrt{x^4+4x}}{2x},\frac{-x^2\mp \sqrt{x^4+4x}}{2x}\right).$