Equilibirium point of the autonomous system ODE What is the equilibrium points of the system $$x'=yz-x^2 \quad , \quad y'=zx-y^2 \quad  , \quad z'=xy-x^2$$
I also wondering how we can determine the trajectories of this system? can we somehow sketch them or imagine them? 
Obviously the origin is one of the Equilibrium points. How we can find others? Is there any specific strategy for finding Equilibrium points?  
 A: How can we find other equilibrium points: Set $$x'=y'=z'=0$$ and then solve for $x,y,z$. This is the typical strategy. In this case start with $z'=x(y-z)=0$ i.e. $x=0$ or $y=z$, then plug this in the other two equations and so on.
Here you get the equilibrium points $(0,0,c)$ with $c \in \mathbb{R}$ and $(a,a,a)$ with $a \in \mathbb{R}$. So you have a whole range of equilibria; the whole $z$-axis consists of equilibria. 
How to sketch the trajectories: You can either compute the solution (normally pretty hard) or you investigate the stability of your equilibria. You can look at Lyapunov functions to determine stability or look at the linearized system around an equilibrium $(x_0,y_0,z_0)$, i.e. $\xi'=J(x_0,y_0,z_0)\xi$ where
$$J(x,y,z)=\begin{pmatrix} -2x & z & y \\ z & -2y & x \\ y-2x &x &0 \end{pmatrix}.$$
Then plug in you equilibria and look for the eigenvalues. We call an equilibrium $(x_0,y_0,z_0)$ stable if $\text{Re}(\lambda)<0$ for all eigenvalues $\lambda$ of $J(x_0,y_0,z_0)$. Unstable if $\text{Re}(\lambda)>0$ for one eigenvalue. And if $\text{Re}(\lambda)=0$ for one eigenvalue then you can't say anything about stability with this method - one also says this is a non-hyperbolic equilibrium and this method (with the Grobman-Hartman-theorem) fails.
Now, for the eigenvalue $\lambda_i$ you can compute the according eigenvector $v_i$. If $\text{Re}\lambda_i<0$ then $v_i$ belongs to the stable eigenspace and if $\text{Re}\lambda_i>0$ then $v_i$ belongs to the unstable eigenspace.
Let me show you an example. You have the equilibrium $(0,0)$ and $\lambda_1=-1, \lambda_2=1, v_1=(1,0), v_2=(0,1)$ then your phase portrait looks like this:

You can see that the trajectories move according to the stable $\langle \binom{1}{0} \rangle$ and unstable $\langle \binom{0}{1} \rangle$ eigenspaces.
Since you liked this method I have drawn some other cases. They all have the same characteristic. First you look at the eigenspaces, draw them and then you draw your orbits according to them. Let's always take wlog $v_1=(1,0), v_2=(0,1)$ since otherwise just rotate your plane.


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*Upper left: $\lambda_1=\lambda_2>0$, $(0,0)$ is unstable

*Upper right: $\lambda_1=\lambda_2<0$, $(0,0)$ is stable

*Lower left: $0<\lambda_1<\lambda_2$, $(0,0)$ is unstable. Since $\lambda_1$ is nearer on the 'critical' zero and is in some sense 'less unstable' than $\lambda_2$, therefore the orbits move more towards $v_2$ 

*Lower right: $0>\lambda_1>\lambda_2$, $(0,0)$ is stable




