Multiplying with $y, z, x$ respectively in the equations we obtain that $x^4y+y^4z$ and cyclic terms are all equal to $xyz$. Thus \[x^4y = y^4z = z^4x = \frac{xyz}{2}\]If one of $x, y, z$ is zero, then it is clear that all of them are zero giving $(0,0,0)$ as a valid triple. Henceforth we assume all of $x, y, z \ne 0$. Then $z = 2x^3$. Using similar identities we obtain that $z = 2x^3 = 2(2y^3)^3 = 16y^9 = 2^{13} z^{27}$. Thus $z = \pm \frac{1}{\sqrt{2}}$. Since a similar condition holds for $x, y$ as well, we obtain that $x^4 = y^4 = z^4$ which gives us that $y = z = x$. Thus $x = y = z = \frac{1}{\sqrt{2}}$ or $x = y = z = -\frac{1}{\sqrt{2}}$. Both of these work, so we have found all solutions, namely $x = y = z = k$ where $k \in \left\{-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right\}$.

Multiplying with $y, z, x$ respectively in the equations we obtain that $x^4y+y^4z$ and cyclic terms are all equal to $xyz$. Thus \[x^4y = y^4z = z^4x = \frac{xyz}{2}\]If one of $x, y, z$ is zero, then it is clear that all of them are zero giving $(0,0,0)$ as a valid triple. Henceforth we assume all of $x, y, z \ne 0$. Then $z = 2x^3$. Using similar identities we obtain that $z = 2x^3 = 2(2y^3)^3 = 16y^9 = 2^{13} z^{27}$. Thus $z = \pm \frac{1}{\sqrt{2}}$. Since a similar condition holds for $x, y$ as well, we obtain that $x^4 = y^4 = z^4$ which gives us that $y = z = x$. Thus $x = y = z = \frac{1}{\sqrt{2}}$ or $x = y = z = -\frac{1}{\sqrt{2}}$. Both of these work, so we have found all solutions, namely $x = y = z = k$ where $k \in \left\{-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right\}$.