= Solution
For $\operatorname{Re}s>1$, the <Euler product> theorem is
$$
\boxed{F(s)=\prod_{p}\left(\sum_{e=0}^\infty f(p^e)p^{-es}\right).}
$$
The local series converge absolutely, and the product is <absolutely convergent> in the sense that the sum over <prime numbers>[primes] of the absolute deviations of the local factors from one is finite. Indeed this sum is bounded by $\sum_{n\ge2}|f(n)|n^{-\operatorname{Re}s}$. To prove the identity, expand the product over <prime numbers>[primes] at most $y$. <Absolute convergence> permits all those local expansions, and multiplicativity identifies the <coefficients>[coefficients] with $f(n)$ for <integers> whose <prime number>[prime] factors are at most $y$. Every omitted <integer> exceeds $y$, so the difference from the full <Dirichlet series> is bounded by its absolute tail beyond $y$, which tends to zero. The same tail bound on a closed right half-plane inside $\operatorname{Re}s>1$ proves local <uniform convergence>. With signed <coefficients>[coefficients] a local factor can be zero; the identity does not require claiming every such general <Euler product> is nonzero.
For $d(n^2)$, multiplicativity gives local <coefficients>[coefficients] $d(p^{2e})=2e+1$. With $z=p^{-s}$,
$$
\sum_{e\ge0}(2e+1)z^e=\frac{1+z}{(1-z)^2}=\frac{1-z^2}{(1-z)^3}.
$$
The divisor bound from (1), applied with a sufficiently small exponent, proves <absolute convergence> when $\operatorname{Re}s>1$; divergence at one follows from $d(n^2)\ge1$. Multiplying the local identities therefore gives
$$
\boxed{\sum_{n\ge1}\frac{d(n^2)}{n^s}=\frac{\zeta(s)^3}{\zeta(2s)}.}
$$
For $n\le x$, the inequality $n^\delta\le x^\delta$ gives $d(n^2)/n\le x^\delta d(n^2)/n^{1+\delta}$. Sum those nonnegative terms and enlarge to the whole infinite series to get the required comparison.
For real $1<s<2$, integral comparison gives $\zeta(s)\le1+(s-1)^{-1}\le2/(s-1)$, while $\zeta(2s)\ge1$. Hence $\zeta(s)^3/\zeta(2s)=O((s-1)^{-3})$ uniformly in that interval. Choosing $\delta=1/\log x$ for $x\ge3$ makes $0<\delta<1$ and $x^\delta=e$. Thus
$$
\boxed{\sum_{n\le x}\frac{d(n^2)}n=O((\log x)^3).}
$$
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