Answer
See below
Work Step by Step
We obtain:
$x-y^2=0\\
y(9x-4)=0$
The critical point are $(0,0);(\frac{4}{9},\frac{2}{3};(\frac{4}{9},-\frac{2}{3})$
The Jacobian of the system is:
$J(x,y)=\begin{pmatrix}
1& -2y\\
9y & 9x-4
\end{pmatrix}$
Substituting:
$J(0,0)=\begin{pmatrix}
1 & 0\\
0 & -4
\end{pmatrix}$
Then the eigenvalues are $\lambda_1=-4,\lambda_2=1$.
Consequently, the equilibrium point $(0,0)$ is a saddle point.
Substituting:
$J(\frac{4}{9},\frac{2}{3})=\begin{pmatrix}
1 & -\frac{4}{3}\\
6 &0
\end{pmatrix}$
Then the eigenvalues are $\lambda_1=\frac{1}{2}+\frac{i\sqrt 31}{2},\lambda_2=\frac{1}{2}-\frac{i\sqrt 31}{2}$.
Consequently, the equilibrium point $(\frac{4}{9},\frac{2}{3})$ is a spiral.
Substituting:
$J(\frac{4}{9},-\frac{2}{3})=\begin{pmatrix}
1 & \frac{4}{3}\\
-6 &0
\end{pmatrix}$
Then the eigenvalues are $\lambda_1=\frac{1}{2}+\frac{i\sqrt 31}{2},\lambda_2=\frac{1}{2}-\frac{i\sqrt 31}{2}$.
Consequently, the equilibrium point $(\frac{4}{9},-\frac{2}{3})$ is a spiral.
