If $|z-\frac 3z|=2$, Find the greatest value of $|z|$ $$|z^2-3|=2|z|$$
And $$|z^2-3|\le |z|^2+3$$
$$2|z|\le |z|^2+3$$ which isn’t a valid equation, what’s going wrong?

Maximizing $a^2+b^2$ with the constraint $(a^2+b^2)^2-10a^2+2b^2+9 = 0$ (please see comments above) results in the two solutions $(3,0)$ and $(-3,0)$, thus the greatest value of $z$ is 3 and occurs at $z\in \{ -3,3\}$.
Here is a picture of the curve plotted in Mathematica along with the circle $|z|=3$.

