You can easily figure out f

First we want to find f(x) from

$2f(x)+f(-x)=\frac{1}{x} \times \sin(x-\frac{1}{x})$

Then substitutes x to be -x and we get

$2f(-x)+f(x)= \frac{1}{x} \times \sin(x-\frac{1}{x})$

Finally we get

$f(x)=\frac{1}{3x} \times \ sin(x-\frac{1}{x})$

So, we substitutes f(x) into the integral

$\int_{1/e}^{e}\frac{1}{x} \times \sin(x-\frac{1}{x})dx =\int_{1/e}^{1}\frac{1}{x} \times \sin(x-\frac{1}{x})dx+\int_{1}^{e}\frac{1}{x} \times \sin(x-\frac{1}{x})dx$

Maybe now someone can think what I'll do next555

We substitutes x to 1/x . Let's we consider this

$\int_{1/e}^{1}\frac{1}{x} \times \sin(x-\frac{1}{x})dx=\int_{e}^{1}x(-\sin(x-\frac{1}{x}))\frac{-1}{x^{2}}dx=-\int_{1}^{e}\frac{1}{x} \times \sin(x-\frac{1}{x})dx$

So, you can see that these two integral cancel to each other then the value is zero!!!

ps1.you can be neglect constant coefficient because since the value is zero it still be zero forever!