Найдите наименьшее значение функции

Решите задачу:

$y=x\, ln^2x\; ,\; \; \; x\in [\, \frac{1}{e^3},e\, ]\\\\y'=ln^2x+2x\, lnx\cdot \frac{1}{x}=lnx(lnx+2)=0\\\\lnx=0\; \; \to \; \; x=1\\\\lnx=-2\; \; \to \; \; x=e^{-2}=\frac{1}{e^2}$

$y(\frac{1}{e^3})=\frac{ln^2\, e^{-3}}{e^3}=\frac{(ln\, e^{-3})^2}{e^3}=\frac{(-3)^2}{e^3}=\frac{9}{e^3}\approx 0,45\\\\y( \frac{1}{e^2} )= \frac{ln^2e^{-2}}{e^2} =\frac{(ln\, e^{-2})^2}{e^2}= \frac{(-2)^2}{e^2} = \frac{4}{e^2}\approx 0,54\\\\y(1)= \frac{ln^21}{1}=0\\\\y(e)=e\cdot ln^2e=e\cdot 1^2=e\approx 2,71828\\\\\\y_{naimen.}=y(1)=0\\\\y_{naibol.}=y(e)=e$