(a)The critical points are solns of these questions.

$$y+x(1-x^2-y^2) = 0$$ $$-x+y(1-x^2-y^2)=0$$

Multiply the first equation by y and the second equation by x . The difference of the two equations gives $x^2+y^2=0$. Hence the only critical point is at the origin.

(b,c) With $F(x,y) = y+x(1-x^2-y^2)$ and $G(x,y) = -x+y(1-x^2-y^2)$, the Jacobian matrix of the vector field is

$$ J = \begin{pmatrix} 1-3x^2-y^2 & 1-2xy \\ -1-2xy & 1-x^2-3y^2 \end{pmatrix} $$

At the origin, the coefficient matrix of the linearized system is

$$ J(0,0) = \begin{pmatrix} 1 & 1 \\ -1 & 1\end{pmatrix} $$

with complex conjugate eigenvalues $r_{1} = 1+i$, $r_{2} = 1-i$, . Hence the origin is an unstable spiral.

Attached is the part (d)