Find all functions $f$ such that $f\left(x^2-y^2\right)=(x-y)\big(f(x)+f(y)\big)$. 
Find all functions $f:\mathbb R\to\mathbb R$ such that
$$f\left(x^2-y^2\right)=(x-y)\big(f(x)+f(y)\big)\text.$$

I have derived these clues:

*

*$f(0)=0$;

*$f(x^2)=xf(x)$;

*$f(x)=-f(-x)$.

But now I am confused. I know solution will be $f(x)=x$, but I don't know how to prove this.

I have found a proof of $f(x)=xf(1)$ without assuming continuity, but only for $x$ integer.
First, let $u=y+x$, $v=y-x$. Then
$$f(uv)=v\left[f\left(\frac{u+v}2\right)+f\left(\frac{u-v}2\right)\right]$$
Now let $u=1$, $v=t$ to get
$$f(t)=t\left[f\left(\frac{t+1}2\right)-f\left(\frac{t-1}2\right)\right]$$
and $u=t$, $v=1$ yields
$$f(t)=\left[f\left(\frac{t+1}2\right)+f\left(\frac{t-1}2\right)\right]$$
Combining these equations it's not difficult to obtain
$$f\left(\frac{t+1}2\right)=f\left(\frac{t-1}2\right)\frac{t+1}{t-1}$$
For $t$ an odd number $\geq3$, write $n=(t-1)/2$ and
$$f(n)=f(n-1)\frac{n}{n-1}$$
Apply recursively this formula and
$$f(n)=nf(1)$$
