Answer
Diverges.
Work Step by Step
Since $$ \int_{0}^{1}\frac{\ln x}{x^2}dx =\int_{0}^{1/3}\frac{\ln x}{x^2}dx +\int_{1/3}^{1}\frac{\ln x}{x^2}dx $$ On $(0,1 / 3]$
$$\ln x\lt-1\ \ \Rightarrow\ \ \ \frac{\ln x}{x^{2}}\lt-\frac{1}{x^{2}}$$
Since
\begin{align*}
\int_{0}^{1/3}\frac{-1}{x^2}dx&= \frac{1}{x}\bigg|_{0}^{1/3}\\
&=-\infty
\end{align*}
Then
$$ \int_{0}^{1}\frac{\ln x}{x^2}dx$$
also diverges.