The solution of the differential equation $$\frac{dy}{dx} = -\left(\frac{x^2+3y^2}{3x^2+y^2}\right)$$, $$y(1) = 0$$ is

Given: $$\frac{dy}{dx} = -\frac{x^2 + 3y^2}{3x^2 + y^2}$$, $$y(1) = 0$$.

Let $$y = vx$$, so $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$.

$$v + x\frac{dv}{dx} = -\frac{x^2 + 3v^2x^2}{3x^2 + v^2x^2} = -\frac{1 + 3v^2}{3 + v^2}$$

$$x\frac{dv}{dx} = -\frac{1 + 3v^2}{3 + v^2} - v = -\frac{1 + 3v^2 + 3v + v^3}{3 + v^2}$$

$$= -\frac{v^3 + 3v^2 + 3v + 1}{3 + v^2} = -\frac{(v+1)^3}{v^2 + 3}$$

$$\frac{v^2 + 3}{(v+1)^3}\, dv = -\frac{dx}{x}$$

Let $$u = v + 1$$, so $$v = u - 1$$, $$v^2 + 3 = u^2 - 2u + 4$$:

$$\frac{u^2 - 2u + 4}{u^3}\, du = -\frac{dx}{x}$$

$$\left(\frac{1}{u} - \frac{2}{u^2} + \frac{4}{u^3}\right)du = -\frac{dx}{x}$$

Integrating:

$$\ln|u| + \frac{2}{u} - \frac{2}{u^2} = -\ln|x| + C$$

$$\ln|u| + \ln|x| + \frac{2}{u} - \frac{2}{u^2} = C$$

$$\ln|ux| + \frac{2}{u} - \frac{2}{u^2} = C$$

Since $$u = v + 1 = \frac{y}{x} + 1 = \frac{x+y}{x}$$:

$$ux = x + y$$

$$\ln|x+y| + \frac{2x}{x+y} - \frac{2x^2}{(x+y)^2} = C$$

$$\ln|1| + \frac{2}{1} - \frac{2}{1} = C \Rightarrow C = 0$$

So: $$\ln|x+y| + \frac{2x}{x+y} - \frac{2x^2}{(x+y)^2} = 0$$

Note that $$\frac{2x}{x+y} - \frac{2x^2}{(x+y)^2} = \frac{2x(x+y) - 2x^2}{(x+y)^2} = \frac{2xy}{(x+y)^2}$$

Therefore: $$\ln|x+y| + \frac{2xy}{(x+y)^2} = 0$$

This matches $$\log_e|x+y| + \frac{2xy}{(x+y)^2} = 0$$.

The correct answer is Option (3): $$\boxed{\log_e|x+y| + \frac{2xy}{(x+y)^2} = 0}$$.