jmc

If $f(n)=3$, this implies that $n=2m^2$ for some positive integer $m$ since the only time $f(n)$ can be odd is when there is an ordered pair $(m,m)$ that cannot be reversed. We start testing values of $m$. The values $m=1$, $m=2$, $m=3$, and $m=4$ do not give $f(n)=3$. However, when $m=5$, we get $50=5^2+5^2=1^2+7^2=7^2+1^2$. Therefore, the smallest integer $n$ for which $f(n)=3$ is $\boxed{50}$.