Delta Epsilon Proof of $\lim_{x\to\infty}\frac{\ln{x}^2}{x}=0$

$\ln x<2\sqrt{x}$ so $$\left|\dfrac{\ln x^2}{x}\right|<\dfrac{4}{\sqrt{x}}<\dfrac{4}{\sqrt{M}}\leq\varepsilon$$ where $x>M$.

... I have started the proof like this:

$\forall \epsilon>0, \exists \delta$ so if $\delta>0$, $|\frac{\ln{x}^2}{x}-0|<\epsilon$.

The $\delta$ in your attempt has nothing to do with $\epsilon$.

By definition, given $\epsilon>0$, you need to show that there exists $M>0$ such that $$ x>M \quad \textbf{implies }\quad |\frac{2\ln x}{x}|<\epsilon\tag{1} $$

Now for $x>1$, $$ \ln x=\int_1^x\frac{1}{t}\ dt\leq \int_1^x\frac{1}{\sqrt{t}}\ dt=2\sqrt{t}\big|_1^x < 2\sqrt{x}, $$ which implies that $$ \frac{2\ln x}{x}<\frac{4\sqrt{x}}{x}=\frac{4}{\sqrt{x}},\quad x>1. $$ So you want $ \frac{4}{\sqrt{x}}<\epsilon, $ which is true if $$ x>(4/\epsilon)^2. $$ Now, setting $M=(4/\epsilon)^2$, you have the implication (1).