$\frac{d}{d x}\left(\int\limits_{x^2}^{x^3} \frac{1}{\log t} d t\right)$ is equal to

$\frac{x^2-x}{\log x}$

Using Leibnitz's rule we have,

$\frac{d}{d x}\left(\int\limits_{x^2}^{x^3} \frac{1}{\log t} d t\right)=\left\{\frac{d}{d x}\left(x^3\right)\right\}\left(\frac{1}{\log x^3}\right)-\left\{\frac{d}{d x}\left(x^2\right)\right\}\left(\frac{1}{\log x^2}\right)$

$\Rightarrow \frac{d}{d x}\left(\int\limits_{x^2}^{x^3} \frac{1}{\log t} d t\right)=\frac{3 x^2}{3 \log x}-\frac{2 x}{2 \log x}=\frac{1}{\log x}\left(x^2-x\right)$