Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{2}-y^{2}\right)=x f(x)-y f(y)
$$
for all pairs of real numbers $x$ and $y$.
The answer is $f(x)=c x, c \in \mathbb{R}$ (these obviously work).
First, by putting $x=0$ and $y=0$ respectively we have
$$
f\left(x^{2}\right)=x f(x) \text { and } \quad f\left(-y^{2}\right)=-y f(y) .
$$

From this we deduce that $f$ is odd, in particular $f(0)=0$. Then, we can rewrite the given as $f\left(x^{2}-y^{2}\right)+f\left(y^{2}\right)=f\left(x^{2}\right)$. Combined with the fact that $f$ is odd, we deduce that $f$ is additive (i.e. $f(a+b)=f(a)+f(b)$ ).
Remark (Philosophy). At this point we have $f\left(x^{2}\right) \equiv x f(x)$ and $f$ additive, and everything we have including the given equation is a direct corollary of these two. So it makes sense to only focus on these two conditions.
Then
$$
\begin{aligned}
f\left((x+1)^{2}\right) & =(x+1) f(x+1) \\
\Longrightarrow f\left(x^{2}\right)+2 f(x)+f(1) & =(x+1) f(x)+(x+1) f(1)
\end{aligned}
$$
which readily gives $f(x)=f(1) x$.