(a) The critical points are solutions of the equations,$1-y = 0$, $(x-y)(x+y) = 0$
The first equation requires that y= 1 . Based on the second equation, x =1/-1. Hence the critical points are (-1,1) and (1,1).
(b,c) $F(x,y) = 1-y$ and $G(x,y) = x^2-y^2 $. The Jacobian matrix of the vector field is $$J = \begin{pmatrix} 0 & -1 \\ 2x & 2y \end{pmatrix}$$
At the critical point $(-1,1)$, the coefficient matrix of the linearized system is $$J(-1,1) = \begin{pmatrix} 0 & -1 \\ -2 & -2 \end{pmatrix}$$
with eigenvalues $r_1 = -1-\sqrt{3}$ and $r_2 = -1+\sqrt{3} $ . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is unstable. At the equilibrium point (1,1), the coefficient matrix of the linearized system is $$J = \begin{pmatrix} 0 & -1 \\ 2 & -2 \end{pmatrix}$$ with complex conjugate eigenvalues $r_1 = -1+i$, $r_2 = -1-i$. The critical point is stable spiral, which is asymptotically stable.
Attached is the part(d)
