Answer
See below
Work Step by Step
We obtain:
$2x+5y^2=0\\
y(3-4x)=0$
The critical point are $(0,0)$
The Jacobian of the system is:
$J(x,y)=\begin{pmatrix}
2 & 10y\\
-4y & 3-4x
\end{pmatrix}$
Substituting:
$J(0,0)=\begin{pmatrix}
2 & 0\\
0 & 3
\end{pmatrix}$
Then the eigenvalues are $\lambda_1=2,\lambda_2=2$.
Consequently, the equilibrium point $(0,0)$ is a node.
