It is known that
$$ \Lambda(n)\log n+\sum_{d|n}\Lambda(d)\Lambda\left(\frac nd\right)=\sum_{d|n}\mu(d)\log^2\frac nd $$
summing this on both side gives
$$ \psi(x)\log x+\sum_{n\le x}\Lambda(n)\psi\left(\frac xn\right)=\sum_{d\le x}\mu(d)\sum_{q\le x/d}\log^2q+\mathcal O(x) $$
It follows from Selberg's asymptotic formula that $RHS=2x\log x+\mathcal O(x)$, so I proceeded with the Mobius inversion identity that
$$ 2x\log x=2\sum_{d\le x}\mu(d)\sum_{q\le x/d}{x\over qd}\log{x\over qd} $$
Plugging this into the RHS, we have
$$ RHS=\sum_{d\le x}\mu(d)\sum_{q\le x/d}\left[\log^2q-{2x\over qd}\log{x\over qd}\right]+2x\log x+\mathcal O(x) $$
Hence, it remains to show that this gigantic double sum is $\mathcal O(x)$, but this is where I became stuck.
NOTE: I am aware that there is an alternative and much simpler proof of this asymptotic formula, but I want to try and see if the method presented in this post is practicable.