Let $f(n)$ be a multiplicative function that is **not** completely multiplicative, i.e $f(m)\cdot f(n)= f(m\cdot n)$ only if $gcd(m,n)=1$. Let $S(x)$ be the double sum over $f$, that is:

$$S(x)=\sum_{i=1}^x\sum_{j=1}^xf(i\cdot j)$$

It is not difficult to see that if $f(n)$ were completely multiplicative, then $S(x)$ could be simplified:

$$S(x)=\sum_{i=1}^x\sum_{j=1}^xf(i\cdot j)= \sum_{i=1}^xf(i)\sum_{j=1}^xf(j)= \biggl(\sum_{k=1}^xf(k)\biggr)^2$$

But since $f(n)$ is not completely multiplicative, this simplification is not completely true, and it fails in every combination where $gcd(i,j)\neq1$. Hence, $S(x)$ can be written this way provided we add some additional error term, let's call it $E$:

$$S(x)=\sum_{i=1}^x\sum_{j=1}^xf(i\cdot j)= \biggl(\sum_{k=1}^xf(k)\biggr)^2+E$$

$E$ is either negative or positive, I'm not sure. Obviously, $E$ is comprised of all the small errors generated by the initial sum term, when $gcd(i,j)\neq1$. I am mainly interested in the cases where $f(n)$ takes the form of:

- Euler totient function: $$S_{\varphi}(x)=\sum_{i=1}^x\sum_{j=1}^x\varphi(i\cdot j)$$
- Sum of divisors function: $$S_{\sigma_1}(x)=\sum_{i=1}^x\sum_{j=1}^x\sigma_1(i\cdot j)$$
- Moebius function: $$S_{\mu}(x)=\sum_{i=1}^x\sum_{j=1}^x\mu(i\cdot j)$$

My question is, what is this error term $E$ exactly? how can I calculate it? How can I properly sum all those small errors to get a correct evaluation of $S(x)$? For clarification, I am concerned with evaluating $S(x)$, but I think I must evaluate $E$ first in order to do it. I am taking this approach because I can compute $\biggl(\sum_{k=1}^xf(k)\biggr)^2$ very efficiently, and so, finding the error term $E$ will solve my question.