= 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 number>[prime] power $p^e$, the divisors are exactly $1,p,\ldots,p^e$, so there are $e+1$. Multiplicativity therefore gives
$$
\boxed{d(n)=\prod_{j=1}^k(e_j+1).}
$$
Fix $\varepsilon>0$. If $p\ge2^{1/\varepsilon}$, then $(e+1)p^{-\varepsilon e}\le(e+1)2^{-e}\le1$ for every <integer> $e\ge0$. There are only finitely many <prime numbers>[primes] below the threshold. For each such <prime number>[prime] the sequence $(e+1)p^{-\varepsilon e}$ tends to zero, so its supremum $C_p$ is finite and at least one. Thus
$$
\frac{d(n)}{n^\varepsilon}=\prod_{p^e\parallel n}(e+1)p^{-\varepsilon e}\le\prod_{p<2^{1/\varepsilon}}C_p=:C(\varepsilon).
$$
The product is over a fixed finite set, independent of $n$, proving $\boxed{d(n)\le C(\varepsilon)n^\varepsilon}$.
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> $K\ge1$, monotonicity of $1/x$ gives $\sum_{k=2}^K1/k\le\int_1^Kdx/x=\log K$; the same bound for real $K\ge1$ follows by taking its <integer> part. Consequently
$$
\frac{\sigma(m)}m\le1+\log d(m)\le\varepsilon\log m+1+\log C(\varepsilon).
$$
Hence the requested bound holds with $\boxed{K(\varepsilon)=1+\log C(\varepsilon)}$, including $m=1$.
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