Answer
Diverges
Work Step by Step
We have: $\ln x \lt \ln (\dfrac{1}{e}) \implies \dfrac{\ln x}{x^{2}}\lt-\dfrac{1}{x^{2}}$
so,$ \dfrac{\ln x}{x^{2}}\lt-\dfrac{1}{x^{2}} \lt 0$
By direct comparison test,
$\int_{0}^{1}\dfrac{-dx}{x^2}=\lim\limits_{a \to 0^{+}} \int_a^1 \dfrac{-dx}{x^2}=1-\lim\limits_{a \to 0^{+}}\dfrac{1}{a}=-\infty$
Thus, the limit does not exist, and so, the integral diverges.