There is a well-known question that seeks the asymptotic behaviour of this function, for $x\geq 2$: $$\sum_{n\leq x} \frac{\phi(n)}{n^2}.$$ See, for example, Apostol "Introduction to Analytic Number Theory" question 6 on page 71.
I have found three posts on stackexchange related to this question, namely post A, post B, and post C. Using methods available to someone who has just read Chapter 3 of Apostol's text, the best solution to this question would seem to be apatch's response to post A.
The method boils down to showing the following identity $$\sum_{d>x}\frac{\mu(d)\log d}{d^2} = O\left(\frac{\log x}{x}\right).$$
This is where I get unstuck. The first step is clearly $$\left|\sum_{d>x}\frac{\mu(d)\log d}{d^2}\right| \leq \sum_{d>x}\frac{\log d}{d^2}$$ but then what do you do with what remains?
One solution I found simply jumps from here to $O(\frac{\log x}{x})$, which seems non-trivial. Apatch's solution in post A does the following: $$\sum_{d>x}\frac{\log d}{d^2}=\sum_{d>x}\frac{\log d}{d^\frac{1}{2}}.\frac{1}{d^\frac{3}{2}}<\frac{\log x}{x^\frac{1}{2}}\sum_{d>x}\frac{1}{d^\frac{3}{2}}$$ but I don't see how can you justify the inequality, since $\frac{\log x}{\sqrt{x}}$ only reaches its maximum until around $x\approx 7.39$, which is well above the $x>2$ condition stated in the question.
Eric Naslund's solution to post C states that $$\sum_{d> x}\frac{\mu(d)\log d}{d^2}=O(1/x)$$ but this less elementary result is presumably even harder to derive.
Can anyone identify the "correct" answer, and explain why?