TXĐ: $x>0$Ta có: $y'=ln^2x+2x.lnx.\frac{1}{x}=ln^2x+lnx$
$y'=0\Leftrightarrow x=\frac{1}{e}\vee x=1$
Ta có: $y(\frac{1}{e})=\frac{1}{e};y(1)=0;y(e)=e;y(e^3)=9e^3$
Vậy $\mathop {\max }\limits_{x \in [e;e^3]}y=9e^3$ khi $x=e^3$
$\mathop {\min }\limits_{x \in [e;e^3]}y=0$ khi $x=1$