Note that $f(a,b)=(4a,4b)$, and thus $f^k(a,b)=(4^ka,4^kb)$. Therefore, $f^k=f$ only when $4^ka=4a$ for all $a\in \mathbb{Z}/18\mathbb{Z}$, and $4^kb=4b$ for all $b\in\mathbb{Z}/60\mathbb{Z}$. If this is true for $a=1$ and $b=1$, then it is true for all $a$ and $b$; and certainly the reverse implication is true.
Thus, we want the smallest $k$ such that $4^k\equiv 4\bmod 18$ and $4^k\equiv 4\bmod 60$. To finish, prove that $4^k\equiv \begin{cases} 10\bmod 18 & \text{ if }k\equiv 0\bmod 3,\\ 4\bmod 18&\text{ if }k\equiv 1\bmod 3,\\ 16\bmod 18&\text{ if }k\equiv 2\bmod 3\end{cases}$ and $4^k\equiv \begin{cases} 16\bmod 60 & \text{ if }k\equiv 0\bmod 2,\\ 4\bmod 60&\text{ if }k\equiv 1\bmod 2\end{cases}$ and find the smallest $k\geq 2$ such that $k\equiv 1\bmod 3$ and $k\equiv 1\bmod 2$.