 Squarefree-part bound for smooth numbers

If $S$ is a finite set of prime numbers, every S-smooth number has a unique expression $n=a{b}^2$, where $a$ is squarefree and its prime factors lie in $S$. There are at most $2^{|S|}$ choices for $a$ and at most $\lfloor \sqrt{x}\rfloor$ choices for $b$ when $n\le x$. Consequently $f_S(x)\le 2^{|S|}\sqrt{x}$.