Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x, y$:

$$
f\left(x^{2}-y^{2}\right)=x f(x)-y f(y) .
$$
With $x=y=0, f(0)=0$. With $y=-x, f(x)=-f(-x)$. And for $y=0$, we get $f\left(x^{2}\right)=$ $x f(x)$. These last two identities allow us to obtain $f(a-b)=\sqrt{a} f(\sqrt{a})-\sqrt{b} f(\sqrt{b})=$ $f(a)-f(b)$ for positive $a, b$. Using the oddness of the function, we also get $f(a-b)+$
$f(-a)=f(-b)$ as well as $f(b-a)=f(b)+f(-a)$. These three formulas allow us to show for all $x, y$ (positive or negative):

$$
f(x)+f(y)=f(x+y)
$$

we have fallen back on Cauchy's equation! Now, let $X$ be real, set $x=\frac{X+1}{2}$ and $y=\frac{X-1}{2}$, which gives us, using Cauchy's equation:

$$
\begin{aligned}
& f(X)=\frac{X+1}{2} f\left(\frac{X+1}{2}\right)-\frac{X-1}{2} f\left(\frac{X-1}{2}\right) \\
& f(X)=f\left(\frac{X}{2}\right)+X f\left(\frac{1}{2}\right) \\
& f(X)=X f(1)
\end{aligned}
$$

Thus $f$ is linear. Conversely, such a function is a solution to our problem.