Problem: Determine all functions $f \colon \RR \to \RR$ such that
\[ f(x^2 - y^2) = x f(x) - y f(y) \]
for all pairs of real numbers $x$ and $y$.

Solution: The answer is $f(x) = cx$, $c \in \RR$ (these obviously work).

First, by putting $x=0$ and $y=0$ respectively
we have \[ f(x^2) = xf(x) \quad\text{and}\quad f(-y^2) = -yf(y). \]
From this we deduce that $f$ is odd, in particular $f(0) = 0$.
Then, we can rewrite the given as $f(x^2-y^2) + f(y^2) = f(x^2)$.
Combined with the fact that $f$ is odd, we deduce that $f$ is additive
(i.e. $f(a+b)=f(a)+f(b)$).

\begin{remark*}
  [Philosophy]
  At this point we have $f(x^2) \equiv xf(x)$ 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.
\end{remark*}

Then
\begin{align*}
  f( (x+1)^2 ) &= (x+1)f(x+1) \\
  \implies f(x^2) + 2f(x) + f(1)
  &= (x+1)f(x) + (x+1)f(1)
\end{align*}
which readily gives $f(x) = f(1)x$.