Let's find the rational solutions of the equation

$$
x+y=x^{2}+y^{2}
$$
If $x=0$, then from $y=y^{2}$ we get $y=0,1$ and $(x, y)=(0,0),(0,1)$ are rational solutions. For the case $x \neq 0$, using the notation $y / x=r$, we have $y=r x$. Substituting this and simplifying by $x$,

$$
x=\frac{1+r}{1+r^{2}}
$$

is obtained, and thus

$$
y=\frac{r+r^{2}}{1+r^{2}}
$$

Therefore, if $(x, y)$ is a rational solution and $x \neq 0$, then $r$ is also rational and

$$
(x, y)=\left(\frac{1+r}{1+r^{2}}, \quad \frac{r+r^{2}}{1+r^{2}}\right)
$$

Furthermore, if $r$ is any rational number, we can verify by direct substitution that

$$
\left(\frac{1+r}{1+r^{2}}, \quad \frac{r+r^{2}}{1+r^{2}}\right)
$$

is indeed a solution, and specifically a rational solution, to the equation.