This has been a long time for me and my analysis reference is pretty abstract.
Could anybody show me again how to calculate the vector field lines for $F(x)=(y-z, z-x, x-y)$?
I know I have to make the differential equation and that the solution will be circles (when I plot it), but I forgot exactly which equation to make. I will be able to solve it of course, just reminding me the equation to solve is enough.
You want curves such that the vector field is tangent to the curve at any point. In other words, you want functions $\gamma:\Bbb R\to \Bbb R^3$ such that $\gamma'(t) = F(\gamma(t))$. Separating out the three component functions of $\gamma$, this is a system of homogenous, linear, first order differential equations with constant coefficients. Which is to say, we may write it using matrix multiplication in the following form: $$ \gamma'(t) = \begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}\gamma(t) $$ Systems like this will have solutions of the form $e^{\lambda t}v$, where $\lambda$ is an eigenvalue of the above matrix, and $v$ is a corresponding eigenvector. And of course, any linear combination of such solutions is also a solution.
So we solve for the eigenvectors and eigenvalues of the above matrix. They turn out to be $$ \begin{array}{|c|c|} \hline\text{Eigenvalue}& \text{Eigenvector}\\ \hline 0 & (1, 1, 1)^T\\ i\sqrt3 & (1-i\sqrt3, 1+i\sqrt3, -2)^T\\ -i\sqrt3 & (1+i\sqrt3, 1-\sqrt3, -2)^T\\ \hline \end{array} $$ A keen eye will spot that an eigenvalue of $0$ means the corresponding eigenvector doesn't change as time goes by (in other words, that's the axis of rotation), while a conjugate pair of pure imaginary eigenvalues corresponds to pure rotation.
If you pick any vector $u\in \Bbb R^3$, and decompose it into a linear combination $a\cdot u_1 + b\cdot u_2 + c\cdot u_3$ of these three eigenvectors, then the field line that goes through $u$ will be given by $$ a\cdot u_1 + b\cdot e^{i\sqrt3t}u_2 + c\cdot e^{-i\sqrt3t}u_3 $$ One can, of course, get rid of any and all mention of complex numbers here by replacing it with trigonometry, but I don't think that it makes the solution any nicer.