If for example $x = 0$, we get a system $0 = y z^2 = 2 y^2 z$ which means that one of the unknowns $y$ and $z$ vanishes and the other can be arbitrary. The cases $y = 0$ or $z = 0$ are discussed similarly. Thus we have obtained three groups of solutions $(x, y, z)$ which are formed by triples $(t, 0, 0)$, $(0, t, 0)$ and $(0, 0, t)$ respectively, where $t$ is any real number. Moreover, we have observed that all the other solutions satisfy the condition $x y z \neq 0$, which is supposed to hold in what follows.

Factorizing the equation $x(y^2 + 2z^2) = y(z^2 + 2x^2)$ yields $(2x - y)(z^2 - x y) = 0$. Thus we distinguish two cases (depending on the fact which of the two factors vanishes).

i. $2x - y = 0$. After setting $y = 2x$ the given system is reduced to the only equation
$$2x(2x^2 + z^2) = 9x^2 z,$$
which can be simplified (by dividing $x \neq 0$) to
$$4x^2 + 2z^2 - 9x z = 0 \quad \text{or} \quad (x - 2z)(4x - z) = 0.$$
Thus the case (i) yields exactly two groups of solutions $(2t, 4t, t)$ and $(t, 2t, 4t)$, where $t$ is any real number.

ii. $z^2 - x y = 0$. Substituting $z^2 = x y$ into the given system, we now get the only equation
$$x y (2x + y) = z(x^2 + 2y^2),$$
which is (because of the inequality $x^2 + 2y^2 > 0$) equivalent to
$$z = \frac{x y (2x + y)}{x^2 + 2y^2}.$$
At this moment we have to find when such a $z$ obeys the condition $z^2 = x y$. After direct substitution we get the following condition on the unknowns $x$ and $y$:
$$\frac{x^2 y^2 (2x + y)^2}{(x^2 + 2y^2)^2} = x y.$$
Dividing by $x y \neq 0$ and removing the fraction yields
$$x y (2x + y)^2 = (x^2 + 2y^2)^2 \quad \text{or} \quad (4y - x)(x^3 - y^3) = 0.$$
Thus we conclude that either $x = 4y$, or $x^3 = y^3$, i.e. $x = y$. Returning to the formula for $z$, we obtain $z = 2y$ or $z = x$, according as $x = 4y$ or $x = y$. Consequently, there are two groups of solutions in the case (ii), namely triples $(4t, t, 2t)$ and $(t, t, t)$, where $t$ is any real number.

*Answer.* All the solutions are $(t, 0, 0)$, $(0, t, 0)$, $(0, 0, t)$, $(t, t, t)$, $(4t, t, 2t)$, $(2t, 4t, t)$ and $(t, 2t, 4t)$, where $t$ is any real number.

To avoid unnecessary repetition from the above solution, we will solve the problem under the condition that $x y z \neq 0$.

Dividing both sides of the given equations by $x y z$ we obtain
$$\frac{y}{z} + \frac{2z}{y} = \frac{z}{x} + \frac{2x}{z} = \frac{x}{y} + \frac{2y}{x}, \quad (3)$$
which can be read as a coincidence of values of a function $f(s) = s + 2/s$ in three nonzero points $s_1 = y/z$, $s_2 = z/x$ and $s_3 = x/y$. Thus we first find when $f(s) = f(t)$ for two nonzero real numbers $s$ and $t$. It follows from the identity
$$f(s) - f(t) = s + \frac{2}{s} - t - \frac{2}{t} = \frac{(s-t)(s t - 2)}{s t}$$
that $f(s) = f(t)$ if and only if $s = t$ or $s t = 2$. Consequently, the system (3) holds if and only if the introduced numbers $s_1, s_2, s_3$ possess the following property: $s_i = s_j$ or $s_i s_j = 2$, for any indices $i$ and $j$. However, if there exists a permutation $(i, j, k)$ of $(1, 2, 3)$ such that $s_i s_j = 2$, then the identity $s_i s_j s_k = 1$ implies that $s_k = \frac{1}{2}$ and hence $s_i \in \{\frac{1}{2}, 4\}$ (because $s_i = s_k$ or $s_i s_k = 2$). Thus the assumption $s_i s_j = 2$ leads to the conclusion that $(s_1, s_2, s_3)$ is a permutation of $(\frac{1}{2}, \frac{1}{2}, 4)$. It is easy to check that exactly three such permutations are satisfactory and yield the solutions $(4t, t, 2t)$, $(2t, 4t, t)$ and $(t, 2t, 4t)$ of the given system. In the remaining case when $s_1 = s_2 = s_3$, the identity $s_1 s_2 s_3 = 1$ implies that $s_i = 1$ for each $i$, which yields the solutions $(t, t, t)$.