Answer
See below
Work Step by Step
We obtain:
$x-2y+5xy=0\\
2x+y=0$
The critical point are $(0,0);(\frac{1}{2},-1)$
The Jacobian of the system is:
$J(x,y)=\begin{pmatrix}
5y+1 & 5x-2\\
2 & 1
\end{pmatrix}$
Substituting:
$J(0,0)=\begin{pmatrix}
1& -2\\
2 & 1
\end{pmatrix}$
Then the eigenvalues are $\lambda_1=1+2i,\lambda_2=1-2i$.
Consequently, the equilibrium point $(0,0)$ is a spiral.
Substituting:
$J(\frac{1}{2},-1)=\begin{pmatrix}
-4 & \frac{1}{2}\\
2 & 1
\end{pmatrix}$
Then the eigenvalues are $\lambda_1=-\frac{3}{2}-\frac{\sqrt 29}{2},\lambda_2=-\frac{3}{2}+\frac{\sqrt 29}{2}$.
Consequently, the equilibrium point $(\frac{1}{2},-1)$ is a spiral.
