Let $$y = y(x)$$ be the solution of the differential equation $$(2x \log_e x)\frac{dy}{dx} + 2y = \frac{3}{x}\log_e x$$, $$x > 0$$ and $$y(e^{-1}) = 0$$. Then, $$y(e)$$ is equal to

$$(2x\ln x)y' + 2y = \frac{3\ln x}{x}$$. Divide by $$2x\ln x$$: $$y' + \frac{1}{x\ln x}y = \frac{3}{2x^2}$$.

IF = $$e^{\int dx/(x\ln x)} = e^{\ln(\ln x)} = \ln x$$.

$$y\ln x = \int \frac{3\ln x}{2x^2}dx$$. Let $$u = \ln x$$, $$dv = 3/(2x^2)dx$$: $$= -3\ln x/(2x) + \int 3/(2x^2)dx = -3\ln x/(2x) - 3/(2x) + C$$.

$$y(e^{-1}) = 0$$: $$0\cdot(-1) = -3(-1)/(2e^{-1}\cdot...) + C$$... Let me use $$y\ln x = -\frac{3}{2x}(\ln x + 1) + C$$.

At $$x = e^{-1}$$: $$0 \cdot (-1) = -\frac{3}{2e^{-1}}(-1+1) + C = 0 + C$$. So $$C = 0$$.

$$y = \frac{-3(\ln x + 1)}{2x\ln x}$$. At $$x = e$$: $$y = \frac{-3(1+1)}{2e\cdot 1} = \frac{-6}{2e} = \frac{-3}{e}$$.

The correct answer is Option (1): $$-3/e$$.