**(a)** Solving $x(3x +2y -30)=0$, $y(2y-x-6)=0$ we get cases

\begin{align*}

&x=0,\ y=0 &&\implies A_1=(0,0),\\

&x=0,\ 2y-x-6=0 &&\implies A_2=(0,3)\\

&y=0,\ 3x+2y-30=0 &&\implies A_3=(10,0),\\

&3x+2y-30=0,\ 2y-x-6=0 &&\implies A_4=(6,6).

\end{align*}

**(b)** Linearizations at these points have matrices

\begin{align*}

&A_1&&A_2&&A_3&&A_4\\

&

\begin{pmatrix}

-30 &\ \ 0\\

0 &-6

\end{pmatrix}

&&

\begin{pmatrix}

-24 &\ \ 0\\

-3 &6

\end{pmatrix}

&&

\begin{pmatrix}

30 &20\\

0 &-16

\end{pmatrix}

&&

\begin{pmatrix}

18 &12\\

-6 &12

\end{pmatrix}\\

\text{with eigenvalues}\\

&\{-30,-6\} &&\{-24,6\} && \{30,-16\} && \{15-\sqrt{63}i,15-\sqrt{63}i \}

\end{align*}

correspondingly.

Therefore $A_1$ is a stable node, $A_2$ and $A_3$ are saddles, $A_4$ is an unstable focal point and since left bottom number is $-6<0$ it is clockwise oriented.

Directions are:

$A_1$: $\mathbf{f}_1=(1,0)^T$, $\mathbf{f}_2=(0,1)^T$ ; Since $-30=\lambda_1 <\lambda_2=-6$, all trajectories have an entry directions $\pm \mathbf{f}_2$ (except two, which have entry directions $\pm \mathbf{f}_1$).

$A_2$: $\mathbf{f}_1=(10,1)^T$ ($\lambda_1=-24$)--stable, $\mathbf{f}_2=(0,1)^T$ ($\lambda_2=6$)--unstable.

$A_3$: $\mathbf{f}_1=(1,0)^T$ ($\lambda_1=30$)--unstable, $\mathbf{f}_2=(1,-2.3)^T$ ($\lambda_2=-16$)--stable.

**(c)--(d)** One should observe that either $x=0$ in every point of the trajectory, or in no point; and that $y=0$ in every point of the trajectory, or in no point. It allows us to make a ``skeleton'' of the phase portrait (thick black lines) on the figure; red lines are very approximate

**Remark**: $A_4$ may not look as a focal point on the computer-generated plot but it is. The reason for this disparity is that for one rotation (so angle increases by $2\pi$) the radius increases in $\exp (2\pi 15/\sqrt{63})\approx 130\,000$ times and this is a really big number.