Let $I_1={\int }_1^e{\left(\frac{x}{e}\right)}^{2x}{{\log }_{e}x}_{ }dx$

Let ${\left(\frac{x}{e}\right)}^x=t$

Taking natural logarithm on both sides, we get

$x\text{ln}\frac{x}{e}=\text{ln }t$

$\Rightarrow x\left(\text{ln\hspace{0.17em}}x-1\right)=\text{ln}t .........\left(i\right)$

Differentiating equation $\left(i\right)$ on both the sides, we get

$\text{ln}x-1+x\left(\frac{1}{x}\right)dx=\frac{1}{t}dt$

$\text{⇒ln}xdx=\frac{1}{t}dt$

$\therefore I_1={\int }_{\frac{1}{e}}^1{t}^2\frac{1}{t}dt={\left(\frac{t^2}{2}\right)}_{\frac{1}{e}}^1=\frac{1}{2}-\frac{1}{2e^2}$

and $I_2={\int }_1^e{\left(\frac{e}{x}\right)}^x{\log }_{e}xdx$

Let ${\left(\frac{e}{x}\right)}^x=t$

Taking natural logarithm on both sides, we get

$x\ln \left(\frac{e}{x}\right)=\ln t$

$\Rightarrow x\left(1-\ln x\right)=\ln t .........\left(ii\right)$

Differentiating equation $\left(ii\right)$ both sides, we get

$\Rightarrow \left[x\left(0-\frac{1}{x}\right)+\left(1-\ln x\right).1\right]dx=\frac{dt}{t}$ (using product rule of differentiation)

$\Rightarrow \ln x=-\frac{dt}{t}$

$\therefore I_2={\int }_e^{1}t.\frac{-dt}{t}={\left(-t\right)}_e^1=\left(-1\right)-\left(-e\right)=e-1$

Hence, required integral is $I_1-I_2$

$=\left(\frac{1}{2}-\frac{1}{2e^2}\right)-\left(e-1\right)$

$=\frac{3}{2}-e-\frac{1}{2e^2}$.