The solution curve of the differential equation $$y\frac{dx}{dy} = x(\log_e x - \log_e y + 1), x > 0, y > 0$$ passing through the point $$(e, 1)$$ is

$$y\frac{dx}{dy}=x(\ln x-\ln y+1)=x\ln\frac{ex}{y}$$. Let $$v=x/y$$: $$x=vy$$, $$\frac{dx}{dy}=v+y\frac{dv}{dy}$$.

$$y(v+y\frac{dv}{dy})=vy\ln(ev)$$. $$v+y\frac{dv}{dy}=v\ln(ev)=v(1+\ln v)$$.

$$y\frac{dv}{dy}=v\ln v$$. $$\frac{dv}{v\ln v}=\frac{dy}{y}$$. $$\ln(\ln v)=\ln y+C$$.

$$\ln v=ky$$. At $$(e,1)$$: $$v=e/1=e$$, $$\ln e=k(1)$$, $$k=1$$.

$$\ln(x/y)=y$$.

The answer is Option (3): $$\log_e\frac{x}{y}=y$$.