#### Answer

We can see that the value is approaching $w_3 = -\frac{1}{2}-\frac{\sqrt{3}}{2}i$.
Therefore, the point $~~z_1 = -1-i~~$ should be colored yellow.

#### Work Step by Step

$f(z) = \frac{2z^3+1}{3z^2}$
$z_1 = -1-i$
We can evaluate the function for $z_1$:
$z_2 = f(z_1)$
$z_2 = \frac{2(-1-i)^3+1}{3(-1-i)^2}$
$z_2 = \frac{2(2-2i)+1}{3(2i)}$
$z_2 = (\frac{5-4i}{6i})(\frac{i}{i})$
$z_2 = \frac{4+5i}{-6}$
$z_2 = -0.667 - 0.833i$
We can evaluate the function for $z_2$:
$z_3 = f(z_2)$
$z_3 = \frac{2(-0.667 - 0.833i)^3+1}{3(-0.667 - 0.833i)^2}$
$z_3 = \frac{2(1.092-0.534i)+1}{3(-0.249+1.111i)}$
$z_3 = (\frac{3.184-1.068i}{-0.747+3.333i})(\frac{-0.747-3.333i}{-0.747-3.333i})$
$z_3 = \frac{-5.938-9.814i}{11.667}$
$z_3 = -0.509-0.841i$
We can see that the value is approaching $w_3 = -\frac{1}{2}-\frac{\sqrt{3}}{2}i$.
Therefore, the point $~~z_1 = -1-i~~$ should be colored yellow.