= Solution
By <superposition>, $y(x)=\int_1^eG(x,\xi)f(\xi)\,d\xi$. Split the <integral> at $\xi=x$: the portion $\xi>x$ uses the first branch of the <Green's function>, and the portion $\xi<x$ uses the second. This yields
$$
\boxed{y(x)=x\ln x\int_x^e(\ln\xi-1)f(\xi)\,d\xi+x(\ln x-1)\int_1^x\ln\xi f(\xi)\,d\xi.}
$$
For $f(\xi)=1/\xi$, put $t=\ln x$. The two <integrals> are
$$
\int_x^e\frac{\ln\xi-1}{\xi}\,d\xi=-\frac12-\frac{t^2}2+t,\qquad\int_1^x\frac{\ln\xi}{\xi}\,d\xi=\frac{t^2}2.
$$
Substitution and cancellation give
$$
\boxed{y(x)=\frac x2\bigl[(\ln x)^2-\ln x\bigr].}
$$
Indeed $y(1)=y(e)=0$. For a direct differential check,
$$
y'=\frac12(t^2+t-1),\qquad y''=\frac{2t+1}{2x},
$$
and therefore
$$
y''-\frac{y'}x+\frac y{x^2}=\frac{(2t+1)-(t^2+t-1)+(t^2-t)}{2x}=\frac1x.
$$
The homogeneous uniqueness shown in part (a) makes this the unique solution.
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