#### Answer

The value is approaching $w_2 = -\frac{1}{2}+\frac{\sqrt{3}}{2}i$.
Therefore, the point $z_1 = i$ should be colored blue.

#### Work Step by Step

$f(z) = \frac{2z^3+1}{3z^2}$
$z_1 = i$
We can evaluate the function for $z_1$:
$z_2 = f(z_1)$
$z_2 = \frac{2(i)^3+1}{3(i)^2}$
$z_2 = \frac{1-2i}{-3}$
We can evaluate the function for $z_2$:
$z_3 = f(z_2)$
$z_3 = \frac{2(\frac{1-2i}{-3})^3+1}{3(\frac{1-2i}{-3})^2}$
$z_3 = \frac{2(\frac{-11+2i}{-27})+1}{3(\frac{-3-4i}{9})}$
$z_3 = \frac{(\frac{22-4i}{27})+\frac{27}{27}}{(\frac{-9-12i}{9})}$
$z_3 = \frac{49-4i}{3(-9-12i)}$
$z_3 = (\frac{49-4i}{-27-36i})~(\frac{-27+36i}{-27+36i})$
$z_3 = \frac{-1179+1872i}{2025}$
$z_3 = -0.582+0.924i$
We can see that the value is approaching $w_2 = -\frac{1}{2}+\frac{\sqrt{3}}{2}i$. Therefore, the point $z_1 = i$ should be colored blue.