Long term Behavior of Dynamical System Given the following dynamical system:
$ \dot x = -6x^2+yz+x-1 $
$ \dot y = 4xz-3y^2+y-2 $
$ \dot z = 9xy-2z^2+z-3 $
What can you say about its long term behavior?
Attempt:
First, finding the fixed points.
There is only one real solution to $x'=0$, $y'=0$ and $z'=0$ and this is at the point $(1,2,3)$.  At this point, the eigenvalues of the Jacobian matrix are 
$\lambda = 1,-17  $  Because of the positive eigenvalue, this fixed point is unstable.  
I have also run into fixed points for the above system but they are complex.  Do these complex fixed points have an influence on the dynamics of the system?  
 A: The change of variables $x \to \xi$, $y \to 2 \eta$, $z \to 3 \zeta$ yields the system
\begin{align}
 \dot{\xi} &= f(\xi) + 6 \eta \zeta, \\
 \dot{\eta} &= f(\eta) + 6 \xi \zeta, \tag{1}\\
 \dot{\zeta} &= f(\zeta) + 6 \xi \eta, 
\end{align}
with
\begin{equation}
 f(x) = -6 x^2 +x-1.
\end{equation}
Not only is the vector field of $(1)$ conservative (its curl vanishes), it is also invariant under all permutations of the triple $(\xi,\eta,\zeta)$. 
This type of systems has been studied extensively by Martin Golubitsky. A good source is
M. Golubitsky, I. Stewart, The Symmetry Perspective, Birkhäuser, Basel, 2002. 
Of particular interest is section 3.4 in chapter 3, on rings of cells.
Addition: The symmetry suggests another coordinate change. Note that the linear combination $\xi+\eta+\zeta$ is invariant under permutations of $(\xi,\eta,\zeta)$. Therefore, introducing the (suitably normalised) coordinates
\begin{align}
 X &= \frac{\xi - \eta}{\sqrt{2}},\\
 Y &= \frac{\xi+\eta - 2 \zeta}{\sqrt{6}},\\
 Z &= \frac{\xi+\eta+\zeta}{\sqrt{3}},
\end{align}
yields
\begin{align}
 \dot{X} &= X(1 - 6 \sqrt{3} Z),\\
 \dot{Y} &= Y(1 - 6 \sqrt{3} Z),\\
 \dot{Z} &= Z-\sqrt{3} - 3 \sqrt{3}(X^2+Y^2).
\end{align}
The equation for $Z$ suggests the introduction of polar coordinates $X = R \cos \theta$, $Y = R \sin \theta$, yielding
\begin{align}
 \dot{R} &= R(1 - 6 \sqrt{3} Z),\\
 \dot{\theta} &= 0\\
 \dot{Z} &= Z-\sqrt{3} - 3 \sqrt{3}R^2.
\end{align}
So, the three-dimensional system is reduced to a planar system.
