Solving a functional equation using even and odd properties

I need help solving this functional equations problem.

Find all $\ f : \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y\ \epsilon \ \mathbb{R}$, the two following equations hold:

$f(3x)=f\left(\frac{x+y}{(x+y)^2+1}\right) + f\left(\frac{x-y}{(x-y)^2+1}\right)$

$f(x^2-y^2)=(x+y)f(x-y)+(x-y)f(x+y)$

What I did was using the second equation to get that $f(0)=0$ by setting $x=y=0$.

then setting $x=0, \ y$ arbitrary in the first equation to get:

$0=f(0)=f\left(\frac{y}{y^2+1}\right) + f\left(\frac{-y}{y^2+1}\right)$
$\implies f\left(\frac{y}{y^2+1}\right)=-f\left(\frac{-y}{y^2+1}\right) \implies f$ is odd.

But then I got stuck here, I tried many ways but always end up with a variation

of $f(-y^2)=2yf(-y)=-2yf(y)$

Any suggestions how to go on from here?