This is a follow-up question to: What is known about partial sums involving squares of the totient function?
Using the hint given by Daniel Fischer, I was able to express the sum posed in the question in the following way:
$$\displaystyle\sum_{r \leq x} \Big( \frac{\varphi(r)}{r} \Big)^2 = x \displaystyle\sum_{r \leq x} \frac{\mu(r)}{r^2} \displaystyle\prod_{p | r} \Big(2 - \frac{1}{p} \Big) = x \displaystyle\sum_{r \leq x} \frac{\mu(r)}{r^2} \displaystyle\sum_{d | r} \frac{\mu(d)}{d} 2^{\omega(r/d)}$$
I am interested in the asymptotic behaviour of this sum, and unfortunately, I am stuck once again. Numerical experimentation suggests that:
$$\displaystyle\sum_{r \leq x} \Big( \frac{\varphi(r)}{r} \Big)^2 = \frac{6}{\sqrt{2} \pi^2} x + O(\text{something})$$
Interestingly enough, we know that:
$$\displaystyle\sum_{n = 1}^{\infty} \frac{\mu(n)}{n^2} = \frac{6}{\pi^2}$$
so I'm inclined to believe that the first sum separates nicely, when we split it into a sum from $1$ to $\infty$, minus the sum for all $r > x$.
I am only familiar with very basic analytic number theory, I don't know of any tricks to use to try to compute the product of primes in the first equation, or do something else that would allow for a clean solution. Any hints would be much appreciated!
$$\sum_{n=1}^\infty (\varphi(n)/n)^2 n^{-s}= F(s)\zeta(s)$$ where $$F(s)=\sum_{m\ge 1} a_m m^{-s}= \prod_p (1-p^{-s})(1+\sum_{k\ge 1} \frac{(p-1)^2}{p^2}p^{-sk})$$ $$= \prod_p (1+O(p^{-1})p^{-s}+\sum_{k\ge 2} O(1)p^{-2s})$$ converges absolutely for $\Re(s)>1/2$. Thus $$\sum_{n\le x} (\varphi(n)/n)^2 = \sum_{m\le 1} a_m \lfloor x/m\rfloor = x \sum_{m\le 1} a_m/m+ O(\sum_{m\le x} |a_m|)=xF(1)+O(x^{1/2+\epsilon})$$ I don't think $$F(1)=\prod_p (1-p^{-1})(1+\frac{p-1}{p^2})$$ has a closed-form.