(a)
\begin{equation}
\left\{
\begin{array}{**lr**}
y+x-x^{3}-xy^{2}=0 & \\
-x+y-x^{2}y-y^{3}=0\\
\end{array}
\right.
\end{equation}
\begin{equation}
\left\{
\begin{array}{**lr**}
x^{2}+y^{2}=0
\end{array}
\right.
\end{equation}
\begin{equation}
\left\{
\begin{array}{**lr**}
x=0 & \\
y=0\\
\end{array}
\right.
\end{equation}
Therefore, the only critical point is (0,0)
(b)
The Jacobian matrix of the vector field is:
\begin{align*}
J &= \begin{bmatrix}
1-3x^{2}-y^{2} & 1-2xy \\
-1-2xy & 1-x^{2}-3y^{2}
\end{bmatrix}\\
~\\
J(0,0) &= \begin{bmatrix}
1 & 1 \\
-1 & 1
\end{bmatrix}\\
\end{align*}
(c)
\begin{align*}
For (0,0), let A&= \begin{bmatrix}
1 & 1 \\
-1 & 1
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
1-\lambda & 1 \\
-1 & 1-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda-1)^{2}+1=0\\
~\\
\lambda &= 1 \pm i \\
~\\
Then \ the \ system \ has \ a \ clockwise \ spiral \ outwards \ at \ (0,0) \\
\end{align*}
(d) In the attachment.
