Step 1: Make $z^3$ alone.
From $3\sqrt3\,z^3-i=0$ we get $z^3=\dfrac{i}{3\sqrt3}$. Notice $\dfrac{1}{3\sqrt3}=\left(\dfrac{1}{\sqrt3}\right)^3$, which is handy.
Step 2: Write the right side in angle form.
Since $i=e^{i\pi/2}$, we have $z^3=\left(\dfrac{1}{\sqrt3}\right)^3 e^{i\pi/2}$. Writing in polar form helps us take cube roots cleanly.
Step 3: Take the three cube roots.
By De Moivre's rule the roots are $z=\dfrac{1}{\sqrt3}\,e^{i(\pi/6+2k\pi/3)}$ for $k=0,1,2$.
Step 4: List the roots and spot the special one.
The angles are $\pi/6$, $5\pi/6$, and $3\pi/2$. The root at $3\pi/2$ points straight down, so it is purely imaginary; its real part is zero. The other two have non-zero real parts.
Step 5: Square and add the two chosen roots.
Squaring doubles the angle, so $z_1^2+z_2^2=\dfrac{1}{3}\left(e^{i\pi/3}+e^{i5\pi/3}\right)$.
Step 6: Simplify the sum.
The pair $e^{i\pi/3}+e^{i5\pi/3}=2\cos\dfrac{\pi}{3}=2\times\dfrac12=1$. So the answer is $\dfrac{1}{3}$. \[ \boxed{\frac{1}{3}} \]