(a)

Set x’ = 0 and y’=0

Then we have critical points (0,0), (-1,0)

(b)

J = \begin{bmatrix}2x+1 & 2y \\-y & 1-x \end{bmatrix}

Linear systems are shown with each critical point:

J(0,0) = \begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}

J(-1,0) = \begin{bmatrix}-1 & 0 \\0 & 2 \end{bmatrix}

(c)

Eigenvalues are computed by det(A - tI)= 0

So that

At (0,0): t=1

Critical point is an unstable proper node or spiral point

At (-1,0): t=-1 and 2

Critical point is an unstable saddle point