 Solution

A multiplicative arithmetic function has $f(1)=1$ and satisfies $f(mn)=f(m)f(n)$ whenever $m,n$ are coprime. For a prime power $p^e$, the divisors are exactly $1,p,\dots ,p^e$, so there are $e+1$. Multiplicativity therefore givesFix $\epsilon >0$. If $p\ge 2^{1/\epsilon }$, then $(e+1)p^{-\epsilon e}\le (e+1)2^{-e}\le 1$ for every integer $e\ge 0$. There are only finitely many primes below the threshold. For each such prime the sequence $(e+1)p^{-\epsilon e}$ tends to zero, so its supremum $C_p$ is finite and at least one. ThusThe product is over a fixed finite set, independent of $n$, proving $\boxed{d(n)\le C(\epsilon )n^{\epsilon }}$.

The change of divisor $k=m/d$ gives $\sigma (m)/m={\sum }_{k\mid m}1/k$. List the divisors as $k_1<\cdots <k_{d(m)}$. They are distinct positive integers, so $k_j\ge j$ and $\sigma (m)/m\le {\sum }_{j\le d(m)}1/j$.

For an integer $K\ge 1$, monotonicity of $1/x$ gives ${\sum }_{k=2}^{K}1/k\le {\int }_1^{K}dx/x=\log K$; the same bound for real $K\ge 1$ follows by taking its integer part. ConsequentlyHence the requested bound holds with $\boxed{K(\epsilon )=1+\log C(\epsilon )}$, including $m=1$.