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 don't know how to do this without continuity. If you can show $f(x) \to f(0)$ whenever $x \to 0$, then the original expression yields it. Then all you need is $f(x)$ bounded near zero in order to use $f(x^2) = x f(x)$ to get that the limit exists and is zero.
However, taking continuity for granted, consider your result $f(x^2)=xf(x)$ which can be rewritten as $f(x) = x^{1/2} f(x^{1/2})$ if we let $x^2 \to x$ for positive $x$. Then $f(x^2) = x \cdot x^{1/2} f(x^{1/2})$. But then we can write $f(x^{1/2}) = x^{1/4} f(x^{1/4})$ if we let $x \to x^{1/4}$ in $f(x^2) = xf(x)$ so that $f(x^2) = x \cdot x^{1/2} \cdot x^{1/4} f(x^{1/4})$.
Repeating this $n$ times we get 
$$f(x^2) = x^{\sum_{n=0} 2^{-n}} f(x^{2^{-n}}) = x^{(2-2^{-n})} f(x^{2^{-n}})$$
which holds for $x>0$. Now for $x > 0$, taking the limit as $n \to \infty$ gives
$$f(x^2) = x^2 f(1)$$
using continuity of $f$ or restated
$$f(x) = f(1) x$$
for positive $x$.
But now for negative $x$ use the fact $f(x) = -f(-x)$ so that $f(-x) = -x f(1)$. Letting $\alpha = f(1)$, we say all solutions must be of the form $f(x) = \alpha x$. We easily verify they are all solutions.
