(a)
Set (2+x)(y-x)=0 and (4-x)(y+x)=0
Then we have critical points (0,0), (4,4), (-2,2)
(b)
J = \begin{bmatrix}-2-2x+y & 2+x \\4-2y-2x & 4-x \end{bmatrix}
Linear systems are shown with each critical point:
J(0,0) = \begin{bmatrix}-2 & 2 \\4 & 4 \end{bmatrix}
J(-2,2) = \begin{bmatrix}4 & 0 \\6 & 6 \end{bmatrix}
J(4,4) = \begin{bmatrix}-6 & 6 \\-8 & 0 \end{bmatrix}
(c)
Eigenvalues are computed by det(A - tI)= 0
So that
At (0,0): t=1±√17}
Critical point is a saddle point and it is unstable
At (-2,2): t= 4 and 6
Critical point is an unstable node
At ((4,4): t=-3±√9 i
Critical point is a stable spiral point