Answer
See below
Work Step by Step
We obtain:
$2x+9y^2=0\\
y(3x-2)=0$
The critical point is $(0,0)$
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.
