Cauchy's equation on the rationals

Step 1. $x=y=0$ gives $f(0)=2f(0)$, so $f(0)=0$. Then $0=f(x+(-x))=f(x)+f(-x)$, so $f$ is odd.

Step 2: integers. By induction $f(n)=nf(1)$ for $n\ge0$: $f(n+1)=f(n)+f(1)$. For negative integers use oddness.

Step 3: rationals. Let $x=\frac mn$ with $n\ge1$. Then
$$n\,f\!\left(\frac mn\right)=f\!\left(\underbrace{\frac mn+\dots+\frac mn}_{n}\right)=f(m)=mf(1),$$
hence $f\!\left(\frac mn\right)=\frac mnf(1)$.

Conclusion. $f(x)=cx$ with $c=f(1)$, and conversely every such function satisfies the equation.

Remark. On $\mathbb{R}$ the same argument only gives $f(x)=cx$ for rational $x$: there exist additive functions $\mathbb{R}\to\mathbb{R}$ that are not linear (they are wild and nowhere continuous). One needs an extra assumption (monotone, bounded on an interval, continuous at a point) to conclude.

Lesson. Start every additive equation with $f(0)=0$, oddness, and the integer case; the rationals follow by "multiply by the denominator".