(a)The critical points are given by the solution set of the equations.$$1-xy = 0$$ $$x-y^3=0$$

After multiplying the second equation by y , it follows that $y=1/-1$ . Hence the critical points of the system are at (-1,1) and (-1,-1).

(b,c) Note that $F(x,y) = 1-xy$ and $G(x,y) = x-y^3$ . The Jacobian matrix of the vector field is $$ J = \begin{pmatrix} -y & -x \\ 1 & -3y^2 \end{pmatrix} $$

**At the critical point** (1,1), the coefficient matrix of the linearized system is$$ J(1,1) = \begin{pmatrix} -1 & -1 \\ 1 & -3 \end{pmatrix}$$

with eigenvalues $r_1 = r_2 = -2$ . The eigenvalues are real and equal. It is easy to show that there is only one linearly independent eigenvector. Hence the critical point is a stable improper node.

**At the point** (-1,-1), the coefficient matrix of the linearized system is $$ J(-1,-1) = \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix}$$

with eigenvalues $r_1 = -1+\sqrt{5}$, $r_2 = -1- \sqrt{5}$. The eigenvalues are real, with opposite sign. Hence the critical point of the associated linear system is a saddle, which is unstable.

Attached is the part(d).