If $$y = yx$$ is the solution of the differential equation $$2x^2\frac{dy}{dx} - 2xy + 3y^2 = 0$$ such that $$y(e) = \frac{e}{3}$$, then $$y(1)$$ is equal to

We have the differential equation $$2x^2\frac{dy}{dx} - 2xy + 3y^2 = 0$$ with $$y(e) = \frac{e}{3}$$. Letting $$y = vx$$ so that $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$ and substituting into the equation gives
$$2x^2\left(v + x\frac{dv}{dx}\right) - 2x(vx) + 3(vx)^2 = 0$$
which simplifies to
$$2x^2 v + 2x^3\frac{dv}{dx} - 2vx^2 + 3v^2 x^2 = 0$$
and hence to
$$2x^3\frac{dv}{dx} + 3v^2 x^2 = 0\,. $$

Rearranging this equation by dividing through by $$x^2$$ yields $$2x\frac{dv}{dx} = -3v^2$$, which can be written in separated form as $$\frac{dv}{v^2} = -\frac{3}{2}\frac{dx}{x}\,. $$

Integrating both sides gives $$-\frac{1}{v} = -\frac{3}{2}\ln|x| + C$$ and hence $$\frac{1}{v} = \frac{3}{2}\ln|x| + C_1\,. $$ Since $$v = \frac{y}{x}$$, it follows that $$\frac{x}{y} = \frac{3}{2}\ln|x| + C_1\,. $$

Applying the initial condition $$y(e) = \frac{e}{3}$$ yields
$$\frac{e}{e/3} = 3 = \frac{3}{2}\ln e + C_1 = \frac{3}{2} + C_1\,, $$
so that $$C_1 = \frac{3}{2}\,. $$ Therefore
$$\frac{x}{y} = \frac{3}{2}(\ln x + 1)\,. $$

Finally, for $$x = 1$$ we have
$$\frac{1}{y(1)} = \frac{3}{2}(\ln 1 + 1) = \frac{3}{2}(0 + 1) = \frac{3}{2}\,, $$
so that $$y(1) = \frac{2}{3}\,. $$

The answer is Option B: $$\frac{2}{3}$$.