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?

Your inequality $2|z|\le|z|^2+3$ is very much valid.  For example, $|z|=2$ satisfies it.
The maximum value of $|z|$ that allows this inequality is the maximum root of the corresponding equality $2|z|=|z|^2+3$, which can be solved by the usual methods for quadratic equations to give $|z|=3$.
Thus $|z|=3$ is an upper bound.  To prove that it is the sharp upper bound, meaning the true maximum, simply try $z=3$ itself in your original equation $|z-(3/z)|=2$.
