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