Let $$y = y(x)$$ be the solution of the differential equation $$(x^2 - 3y^2)dx + 3xy$$ dy = 0, $$y(1) = 1$$. Then $$6y^2(e)$$ is equal to

We solve the differential equation $$(x^2 - 3y^2) \, dx + 3xy \, dy = 0$$ with $$y(1) = 1$$.

Rearranging the equation yields $$\frac{dy}{dx} = -\frac{x^2 - 3y^2}{3xy} = \frac{3y^2 - x^2}{3xy}$$, which is a homogeneous equation. Let $$y = vx$$ so that $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$.

Substituting gives $$v + x\frac{dv}{dx} = \frac{3v^2x^2 - x^2}{3x \cdot vx} = \frac{3v^2 - 1}{3v}$$, and hence $$x\frac{dv}{dx} = \frac{3v^2 - 1}{3v} - v = \frac{3v^2 - 1 - 3v^2}{3v} = \frac{-1}{3v}$$.

Separating variables leads to $$3v \, dv = -\frac{dx}{x}$$, which upon integration gives $$\frac{3v^2}{2} = -\ln|x| + C$$.

Substituting back $$v = \frac{y}{x}$$ yields $$\frac{3y^2}{2x^2} = -\ln|x| + C$$. Using the initial condition $$y(1) = 1$$ gives $$\frac{3\cdot 1}{2\cdot 1} = -\ln 1 + C = C$$, so $$C = \frac{3}{2}$$.

Therefore $$\frac{3y^2}{2x^2} = \frac{3}{2} - \ln x$$, which simplifies to $$3y^2 = 3x^2 - 2x^2 \ln x$$ and hence $$y^2 = x^2 - \frac{2x^2 \ln x}{3} = x^2\left(1 - \frac{2\ln x}{3}\right)$$.

At $$x = e$$ we have $$y^2(e) = e^2\left(1 - \frac{2\cdot 1}{3}\right) = e^2 \cdot \frac{1}{3} = \frac{e^2}{3}$$, so $$6y^2(e) = 6 \cdot \frac{e^2}{3} = 2e^2$$.

The answer is $$\boxed{2e^2}$$, which corresponds to Option C.

But the expected answer is 122 (which doesn't match any option label). Our calculation gives Option C: $$2e^2$$. Saving for review.