Step 1: Understand the trick in the question.
We are told $2f(x)-3f\!\left(\frac{1}{x}\right)=x$. We do not yet know $f(x)$ by itself. The smart idea is to make a second equation so we can solve for $f(x)$ alone.
Step 2: Make the second equation.
Replace every $x$ by $\frac{1}{x}$ in the given equation. Note that the partner $\frac{1}{x}$ becomes $x$ again. \[ 2f\!\left(\tfrac{1}{x}\right)-3f(x)=\tfrac{1}{x}. \]
Step 3: Line up the two equations.
Equation 1: $2f(x)-3f\!\left(\frac{1}{x}\right)=x$.
Equation 2: $-3f(x)+2f\!\left(\frac{1}{x}\right)=\frac{1}{x}$.
Step 4: Remove the unwanted term.
Multiply Equation 1 by $2$ and Equation 2 by $3$, then add. This cancels $f\!\left(\frac{1}{x}\right)$. \[ 4f(x)-9f(x)=2x+\tfrac{3}{x} \] \[ -5f(x)=2x+\tfrac{3}{x}. \] So \[ f(x)=-\tfrac{2}{5}x-\tfrac{3}{5x}. \]
Step 5: Set up the integral.
Now we just integrate this simple function from $1$ to $e$. \[ \int_1^e f(x)\,dx=\int_1^e\left(-\tfrac{2}{5}x-\tfrac{3}{5x}\right)dx. \]
Step 6: Integrate term by term.
Using $\int x\,dx=\frac{x^2}{2}$ and $\int\frac{1}{x}\,dx=\ln|x|$, \[ =\left[-\tfrac{1}{5}x^2-\tfrac{3}{5}\ln|x|\right]_1^e. \]
Step 7: Put the limits in.
Remember $\ln e=1$ and $\ln 1=0$. \[ =\left(-\tfrac{e^2}{5}-\tfrac{3}{5}\right)-\left(-\tfrac{1}{5}-0\right)=-\tfrac{e^2}{5}-\tfrac{2}{5}. \]
Step 8: Final value.
This equals $-\frac{2+e^2}{5}$, which is option (1).
\[ \boxed{-\dfrac{2+e^2}{5}} \]