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