Consider the group $G=\mathbb{Z}/3$ with the given representation $\rho:G\rightarrow GL_3(\mathbb{K})$, such that $0\mapsto Id,\ 1\mapsto ((x,y,z)\mapsto (z,x,y)),\ 2\mapsto ((x,y,z)\mapsto(y.z.x))$.
If $\mathbb{K}=\mathbb{R}$ we see that we can split $\mathbb{R}^3$ as $W\oplus V$, with $W=\{(x,y,z)\in \mathbb{R}\mid x+y+z=0\},\ V=\{r(1,1,1)\mid r\in \mathbb{R}\}$. and $W$ cannot be further decomposed. If $\mathbb{K}$ is algebraically closed we can split $W$ further, so let's see the case $\mathbb{K}=\mathbb{C}$, considering the complexification $W\otimes_{\mathbb{R}}\mathbb{C}$. The solution and I tried to argue that way: $W$ is a two dimensional subspace so we want $W_1,W_2$ to be 1-dimensional. $dim_{\mathbb{R}}(W)=2=dim_{\mathbb{C}}(W\otimes\mathbb{C})=dim_{\mathbb{C}}(W_1+W_2))$.
The above two spaces are also given but I don't see how one should get this choice. $W_1=\{ (x,jx,j^2x)\mid x\in \mathbb{C}\},\ W_2=\{(x,j^2x,jx)\mid x\in \mathbb{C}\}$, where $j=e^{2\pi i/3}$. Does it make sense?
Maybe the easiest way to see this is that for the 2-dimensional representation $W\cong \mathbb{R}^2$ in your post, the generator of $G$ acts by a rotation by $2\pi/3$: $$\rho'([1])=\begin{pmatrix}\cos(2\pi/3)&-\sin(2\pi/3)\\\sin(2\pi/3)&\cos(2\pi/3)\end{pmatrix}$$ Here, $\rho':G\to\mathrm{GL}_2(\mathbb{R})$ is the corresponding representation.
This matrix has no eigenvectors in $\mathbb{R}^2$ (the characteristic polynomial of the matrix above does not split), but it has two eigenvectors in $\mathbb{C}^2$ and the corresponding eigenvalues are $j$ and $j^2$ in your notation. This is why $W\otimes\mathbb{C}$ decomposes into a direct sum of two irreducibles over $\mathbb{C}$.
Now, $W=\{(a,b,c)\in\mathbb{R}^3\mid a+b+c=0\}\cong\mathbb{R}^2$, and you know now that you want to find vectors in $W\otimes\mathbb{C}$ on which $[1]$ acts by either $j$ or $j^2$. But also, $[1].(a,b,c)=(c,a,b)$ since this is the natural action of $G$ on $\mathbb{R}^3$. Comparing the two formulas gives you the desired result.
Indeed, the eigenvector with eigenvalue $j$ satisfies $[1].(a,b,c)=(ja,jb,jc)$. So, $$(ja,jb,jc)=(c,a,b)$$ Hence $c=ja$, and $b=jc=j^2a$, so the $j$-eigenspace for the action of $[1]$ is the span of $(1,j^2, j)$. Similarly, the $j^2$-eigenspace is the span of $(1,j,j^2)$.