Let $$x = x(y)$$ be the solution of the differential equation $$2ye^{x/y^2} dx + \left(y^2 - 4xe^{x/y^2}\right) dy = 0$$ such that $$x(1) = 0$$. Then, $$x(e)$$ is equal to

We need to solve $$2ye^{x/y^2}\,dx + (y^2 - 4xe^{x/y^2})\,dy = 0$$ with $$x(1) = 0$$, and find $$x(e)$$.

Rearrange the equation:

Dividing by $$dy$$:

$$2ye^{x/y^2}\frac{dx}{dy} + y^2 - 4xe^{x/y^2} = 0$$

Substitute $$v = \frac{x}{y^2}$$, so $$x = vy^2$$:

Then $$\frac{dx}{dy} = v'y^2 + 2vy$$ where $$v' = \frac{dv}{dy}$$.

Substituting:

$$2ye^v(v'y^2 + 2vy) + y^2 - 4vy^2e^v = 0$$

$$2v'y^3e^v + 4vy^2e^v + y^2 - 4vy^2e^v = 0$$

$$2v'y^3e^v + y^2 = 0$$

Separate variables:

$$2v'y^3e^v = -y^2$$

$$e^v\,dv = -\frac{1}{2y}\,dy$$

Integrate both sides:

$$\int e^v\,dv = -\frac{1}{2}\int \frac{dy}{y}$$

$$e^v = -\frac{1}{2}\ln y + C$$

$$e^{x/y^2} = -\frac{1}{2}\ln y + C$$

Apply initial condition $$x(1) = 0$$:

When $$y = 1, x = 0$$: $$e^0 = -\frac{1}{2}\ln 1 + C$$, so $$C = 1$$.

$$e^{x/y^2} = 1 - \frac{1}{2}\ln y$$

Find $$x(e)$$:

When $$y = e$$:

$$e^{x/e^2} = 1 - \frac{1}{2}\ln e = 1 - \frac{1}{2} = \frac{1}{2}$$

$$\frac{x}{e^2} = \ln\left(\frac{1}{2}\right) = -\ln 2$$

$$x = -e^2\ln 2$$

The correct answer is Option D: $$\boxed{-e^2\log_e 2}$$.