1. Identify the type of differential equation

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

This is a homogeneous differential equation because the right-hand side is a function of $y/x$.

2. Use the substitution

Let $y=vx \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

Substitute into the equation: $v+x\frac{dv}{dx}=-\frac{x^2+3v^2x^2}{3x^2+v^2x^2}=-\frac{1+3v^2}{3+v^2}$

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

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

Factor the numerator: $1+3v+3v^2+v^3=(v+1)^3$

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

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

3. Integrate

We need to evaluate $\int \frac{v^2+3}{(v+1)^3}dv$

Let $u=v+1 \Rightarrow v=u-1$

Then $v^2+3=(u-1)^2+3=u^2-2u+4$

So, $\frac{v^2+3}{(v+1)^3}=\frac{u^2-2u+4}{u^3}=\frac1u-\frac{2}{u^2}+\frac{4}{u^3}$

Thus, $\int \frac{v^2+3}{(v+1)^3}dv=\int\left(\frac1u-\frac{2}{u^2}+\frac{4}{u^3}\right)du$

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

Replacing $u=v+1$: $\ln|v+1|+\frac{2}{v+1}-\frac{2}{(v+1)^2}=-\ln|x|+C$

4. Substitute back $v=\dfrac{y}{x}$

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

we get $\ln\left|\frac{x+y}{x}\right|+\frac{2}{(x+y)/x}-\frac{2}{((x+y)/x)^2}=-\ln|x|+C$

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

The $-\ln|x|$ terms cancel from both sides, so $\ln|x+y|+\frac{2x}{x+y}-\frac{2x^2}{(x+y)^2}=C$

Now simplify the rational part: