Let $f\left(t\right)=\ln \text{t}$ then $\frac{d}{dx}\left\{\underset{x^2}{\overset{x^3}{\int }}f\left(t\right)dt\right\}$

$f\left(t\right)=\ln .t$

$\frac{d}{dx}\underset{x^2}{\overset{x^3}{\int }}f\left(t\right).dt=f\left(x^3\right).3x^2-f\left(x^2\right).2x$

$=\ln x^3\left(3x^2\right)-\ln x^2\left(2x\right)$

$=9x^2\ln x-4x.\ln x$

$Z=x\ln x(9x-4)$

$\frac{dz}{dx}=(1+{\log }_{e}x)(9x-4)+9x.{\log }_{e}x$

$\left(\text{a}\right)x=0\Rightarrow Z=0$

$\left(\text{b}\right)x=1\Rightarrow Z=0;x=\frac{4}{9}\Rightarrow Z=0$

$\left(\text{c}\right)x=e\Rightarrow Z=e.{\log }_{e}e(9e-4)=9e^2-4e$

$\left(\text{d}\right)x=e\Rightarrow \frac{dz}{dx}=(1+1)(9e-4)+9e=18e-8+9e=27e-8$