(a)The critical points are given by the solution set of the equations $$-2x-x-y-x(x^2+y^2) = 0$$ $$x-y+y(x^2+y^2) = 0$$ It is clear that the origin is a critical point. Solving the first equation for C , we find that $$y = \frac{-1 \pm \sqrt{1-8x^2-4x^4}}{2x}$$

Substitution of these relations into the second equation results in two equations of the form $f_{1}(x) = 0$ and $f_{2}(x) = 0$ . Plotting these functions, we note that only$f_{1}(x) = 0$ has real roots. ('It follows that the additional critical points are at (-0.33076,1.0924) and (0.33076,-1.0924)

(b,c) Given that $$F(x,y) = -2x-x-y-x(x^2+y^2)$$ $$G(x,y) = x-y+y(x^2+y^2)$$

the Jacobian matrix of the vector field is $$\begin{pmatrix} -2-3x^2-y^2 & -1-2xy \\ 1+2xy & -1+x^2+3y^2 \end{pmatrix}$$

with complex conjugate eigenvalues $r_{1,2} =((-3 \pm i \sqrt{3})/2) $ . Hence the critical point is a stable spiral, which is asymptotically stable. At the point (-0.33076,1.0924) , the coefficient matrix of the linearized system is $$J(-0.33076,1.0924) = \begin{pmatrix} -3.5216 & -0.27735 \\ 0.27735 & 2.6895 \end{pmatrix} $$

With eigenvalues $r_1 = -3.5092$ and $r_2 = 2.6771$ The eigenvalues are real, with opposite sign. Hence the critical point of the associated linear system is a saddle, which is unstable. Identical results hold for the point at (0.33076,-1.0924).

Attached is the part(d)