(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.
