Answer
See below
Work Step by Step
We obtain:
$2x-9y^2=0\\
y(3x-2)=0$
The critical point are $(0,0);(\frac{2}{3},\frac{2}{3\sqrt 3};(\frac{2}{3},-\frac{2}{3\sqrt 3})$
The Jacobian of the system is:
$J(x,y)=\begin{pmatrix}
3y& 3x-2\\
2 & -18y
\end{pmatrix}$
Substituting:
$J(0,0)=\begin{pmatrix}
0 & -2\\
2 & 0
\end{pmatrix}$
Then the eigenvalues are $\lambda_1=\pm 2i$.
Consequently, the equilibrium point $(0,0)$ is a spiral point.
Substituting:
$J(\frac{2}{3},\frac{2}{3\sqrt 3})=\begin{pmatrix}
\frac{2}{\sqrt 3} & 0\\
2 & -4\sqrt 3
\end{pmatrix}$
Then the eigenvalues are $\lambda_1=-4\sqrt 3,\lambda_2=\frac{2}{\sqrt 3} $.
Consequently, the equilibrium point $(\frac{2}{3},\frac{2}{3\sqrt 3})$ is a saddle point.
Substituting:
$J(\frac{2}{3},-\frac{2}{3\sqrt 3})=\begin{pmatrix}
-\frac{2}{\sqrt 3} & 0\\
2 & 4\sqrt 3
\end{pmatrix}$
Then the eigenvalues are $\lambda_1=4\sqrt 3,\lambda_2=-\frac{2}{\sqrt 3} $.
Consequently, the equilibrium point $(\frac{2}{3},-\frac{2}{3\sqrt 3})$ is a saddle point.
