Answer
$x^2\ln x$ grows faster than $\ln^2x$.
Work Step by Step
We will find the limit $\lim_{x\to\infty}\frac{x^2\ln x}{\ln^2 x}.$
1) If it is equal to zero then $x^2\ln x$ grows slower than $\ln^2 x$;
2) If it is equal to $\infty$ then $x^2\ln x$ grows faster than $\ln^2 x$;
3) If it is equal to some constant then their growth rates are comparable.
$$\lim_{x\to\infty}\frac{x^2\ln x}{\ln^2x }=\lim_{x\to\infty}\frac{x^2}{\ln x}=\left[\frac{\infty^2}{\ln\infty}\right]=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(x^2)'}{(\ln x)'}=\lim_{x\to\infty}\frac{2x}{\frac{1}{x}}=\lim_{x\to\infty}2x^2=\left[2\infty^2\right]=\infty,$$
thus $x^2\ln x$ grows faster than $\ln^2x$.
