More properly speaking, there's a function $f$ over the real numbers whose derivative is known to grow like $\Theta\frac{x^2}{\log x}$. Hence, $f$'s growth rate is something like $\int \frac{x^2}{\log x}$.
Since $x < \frac{x^2}{\log x} < x^2$, I'd suppose that $ \int \frac{x^2}{\log x}$ ($\approx f?$) grows at least as fast as $x^2$ and at most as $x^3$. Or, in other words, $f\in O(x^2)\cap o(x^3)$. But this doesn't rule out something like $f\in\Theta\frac{x^3}{\log x}$.
With some searching, I found that this is related to the "Exponential Integral". But the search results were nearly all about complex analysis and don't help determine the growth rate over positive reals as far as I'm able to tell. There were some weak bounds like $$\text{Ei}(x)\ge \log x + \frac{3}{4}x +(\text{the Euler-Mascheroni constant)}$$ and some convoluted bounds on similar functions (something about branch points?). But whenever I tried "translating" these messier bounds into a growth rate, things fell apart (or the resulting inequality didn't even apply to the positive reals).
We have $\int \frac{x^2}{\log x} \in \Theta(\frac{x^3}{\log x})$ as $x \to \infty$.
To see this, call the antiderivative $F(x)$ (it is really $\operatorname{Ei}(3 \log x)$ but we don't need to know that). Then to prove this result, we want to understand the limit $$\lim_{x \to \infty} \frac{F(x)}{x^3/\log x}.$$ Applying L'Hopital's rule once, we get $$\lim_{x \to \infty} \frac{\frac{x^2}{\log x}}{\frac{3x^2}{\log x} - \frac{x^2}{(\log x)^2}} = \lim_{x \to \infty} \frac{1}{3 - \frac{1}{\log x}} = \frac13.$$ So in fact $F(x) \in \Theta(\frac{x^3}{\log x})$ and more precisely $F(x) \sim \frac{x^3}{3 \log x}$ as $x \to \infty$.
More generally we have $$\int \frac{x^k}{\log x} \sim \frac{x^{k+1}}{(k+1) \log x} \text{ as }x \to \infty$$ by the same argument.