I am not sure if my calculation is correct:
$x^T = [x_1\;\; x_2]^T$ $$\frac{dx_1}{dt} = (6-0.5x_1-3x_2)x_1$$ $$\frac{dx_2}{dt} = (-3-3x_2+x_1)x_2$$
1st equilibrium point:
$$6-0.5x_1-3x_2 = 0\qquad -3-3x_2+x_1=0$$
$$EP_1 = [6 \;\; 1]^T$$
2nd:
$$x_1 = 0 \qquad x_2 = 0 \qquad $$$$EP_2 = [0 \;\; 0]^T$$
3rd:
$$x_1 = 0 \qquad -3-3x_2=0 \qquad $$$$EP_3 = [0 \;\; -1]^T$$
4th:
$$x_2 = 0 \qquad 6-0.5x_1=0 \qquad $$$$EP_4 = [12 \;\; 0]^T$$
I simply divided $x_1$ and $x_2$, then set once $x_1 = 0$, once $x_2 = 0$ and then both zero. $x_1,x_2 = 0$
But it is a new exam, and I don't know if I am correct here.
You have $(6-\frac{1}{2} x_1-3x_2)x_1 = 0$ iff $x_1=0$ or $x_1 \neq 0$ and $-\frac{1}{2} x_1-3x_2 = -6$. Similarly, $(-3-3x_2+x_1)x_2 = 0$ iff $x_2=0$ or $x_2 \neq 0$ and $-3x_2+x_1= 3$.
So, you can find all solutions by taking each of the four cases:
(1) $x_1 = 0$, $x_2 = 0$.
(2) $x_1 = 0$, $x_2 \neq 0$ and $-3x_2+x_1= 3$. This gives $x_2 = -1$.
(3) $x_1 \neq 0$ and $-\frac{1}{2} x_1-3x_2 = -6$, and $x_2 = 0$. This gives $x_1 = 12$.
(4) $x_1 \neq 0$ and $-\frac{1}{2} x_1-3x_2 = -6$, and $x_2 \neq 0$ and $-3x_2+x_1= 3$. This gives $x_1 = 6$. $x_2 = 1$.