Equilibrium points for a system I got the following system:
$\dot x = -x+xy$
$\dot y = x-y -x^{2}-y^{3}$
And I want to find its fixed (equilibrium) points. So, I know I have to put $\dot x = 0$ and $\dot y = 0$, and for $\dot x = 0$ we have that $x(-1+y)=0$, that means, $\dot x = 0$ if $x=0$ and if $y=1$. 
For $\dot y = x-y -x^2-y^3$ is where the problem starts. What is the technique to identify the $0$'s in this case? 
I tried something like $x(1-x^2) = y(1+y^2)$, so the RHS is $0$ only for $y=0$, the LHS is $0$ for $x=1,x=-1,x=0.$ From this and by inspection I got that for $(x,y)=(1,0), (0,0), (-1,-1)$ but this was by inspection. So how can I do this like rigorously? and how many fixed points should I find? 
Any help is welcome, thanks in advance.
 A: We have 
$$\begin{align} \dot x &= -x+x y \\ \dot y &= x-y -x^{2}-y^{3} \end{align}$$
From the first equation, we get
$$-x+x y = 0 \implies x= 0, y = 1$$
For $x = 0$, in the second equation we get
$$y -y^{3} = 0 \implies y = 0, y = \pm i$$
We can only get real roots, so only the root $(x, y) = (0,0)$ counts.
For $y = 1$ in the second equation we get
$$-x^2 -x - 2 = 0 \implies x = \dfrac{1}{2} \left(1 \pm~ i \sqrt{7}\right)$$
Thus no roots from that.
We only have a single critical point.
We can draw a contour plot and see this

A: Easier. You need to solve the system:
$$\begin{cases}
-x+xy = 0\\
x-y -x^{2}-y^{3} = 0
\end{cases}.$$
From the first equation, you get $x = 0$ and $y = 1.$
Using $x = 0,$ the second equation reads as:
$$0-y-0-y^3 = 0\Rightarrow y = 0.$$
Note that the previous equation has also $\pm i$ as solutions. This should be discarded if you are concerned with the study of real-valued dynamical systems.
Therefore, $[0, 0]$ is an equilibrium of your dynamical system.
On the other hand, for $y=1$, the second equation reads as:
$$x-1-x^2-1 = 0\Rightarrow x^2-x+2 = 0.$$ This equation has two complex conjugate solutions, namely:
$$x = \frac{1 \pm i\sqrt{7}}{2},$$
but these solutions are not feasible for real-valued systems.
In conclusion, you system has only one real-valued equilibrium, namely $[0, 0]$. Every nonreal solutions should be discarded.
