Let x = x(y) be the solution of the differential equation 2y^2...

The given differential equation is $2y^2 \dfrac{dx}{dy} - 2xy + x^2 = 0$.

Dividing by $2y^2$, we get:

$\dfrac{dx}{dy} - \dfrac{x}{y} + \dfrac{x^2}{2y^2} = 0$

This is a homogeneous differential equation. Let $x = vy$, then $\dfrac{dx}{dy} = v + y \dfrac{dv}{dy}$.

Substituting these into the equation:

$v + y \dfrac{dv}{dy} - v + \dfrac{v^2}{2} = 0$

$y \dfrac{dv}{dy} = -\dfrac{v^2}{2}$

Separating the variables:

$\dfrac{dv}{v^2} = -\dfrac{dy}{2y}$

Integrating both sides:

$\int v^{-2} dv = -\dfrac{1}{2} \int \dfrac{dy}{y}$

$-\dfrac{1}{v} = -\dfrac{1}{2} \ln y - C$

Substituting $v = \dfrac{x}{y}$:

$\dfrac{y}{x} = \dfrac{1}{2} \ln y + C$

Given $x(e) = e$, substituting $x = e$ and $y = e$:

$\dfrac{e}{e} = \dfrac{1}{2} \ln e + C$

$1 = \dfrac{1}{2} + C \Rightarrow C = \dfrac{1}{2}$

The particular solution is:

$\dfrac{y}{x} = \dfrac{1}{2} \ln y + \dfrac{1}{2} = \dfrac{\ln y + 1}{2}$

$x = \dfrac{2y}{\ln y + 1}$

To find $x(e^2)$, substitute $y = e^2$:

$x(e^2) = \dfrac{2e^2}{\ln(e^2) + 1} = \dfrac{2e^2}{2 + 1} = \dfrac{2}{3} e^2$

Answer: $\dfrac{2}{3} e^2$