= Solution
The initial point lies in $x>1$, where division by $\ln x$ is valid. The <first-order linear differential equation>
$$
y'+\frac{2}{x\ln x}y=\frac1{\ln x}
$$
has <integrating factor> $(\ln x)^2$. Hence
$$
\frac{d}{dx}\bigl[y(\ln x)^2\bigr]=\ln x,\qquad y(\ln x)^2=x\ln x-x+C.
$$
The condition at $x=e$ sets $C=0$, so
$$
\boxed{y(x)=\frac{x(\ln x-1)}{(\ln x)^2},\qquad x>1.}
$$
The solution cannot be continued regularly through $x=1$: its numerator tends to $-1$ while the denominator tends to zero. The real logarithm also requires $x>0$.
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