Finding the asymptotic of the function $\Lambda(x):=\sum_{1 \leq m,n \leq x \,\land \,\gcd(m,n)=1} \frac{1}{mn}$ Inspired by this question Is there a known asymptotic for $A(x):=\sum_{1\leq i,j \leq X} \frac{1}{\text{lcm}(i,j)}$? I tried to find the asymptotic of the following function.
$$
\Lambda(x)=\sum_{\substack{ 1 \leq m,n \leq x \\ \text{gcd}(m,n)=1}} \frac{1}{mn}. 
$$
My approach:
$$
\left(\sum_{1\leq k \leq x} \frac{1}{k}\right)^2=\sum_{1\leq l \leq x} \frac{\Lambda\big(\frac{x}{l}\big)}{l^2}\label{1}\tag{1}
$$
Now,
$$
f(x)=\left(\sum_{1\leq k \leq x} \frac{1}{k}\right)^2≈(\ln(x)+\gamma)^2
$$
From, \eqref{1} can establish the approximate identity
$$
2f(x)-\Lambda(x)≈ 2\int_{1}^{x} \frac{\Lambda(\frac{x}{t})}{t^2} dt
\label{2}\tag{2}$$
or,
$$
2f(x)-\Lambda(x)≈ \frac{2}{x}\int_{1}^{x} {\Lambda(\varphi)} d\varphi 
$$
Using the Newton-Leibniz rule we get
$$x\Lambda'(x)+3\Lambda(x)≈4(\ln(x)+\gamma)+2(\ln(x)+\gamma)^2$$
Solving this differential equation we get,
$$
\Lambda(x)≈\frac{2}{3}\ln^2(x)+\left(\frac{8}{9}+\frac{4}{3}\gamma\right)\ln(x)+\left(\frac{2}{3}\gamma^2+\frac{8}{9}\gamma-\frac{8}{27}\right)+\frac{c_1}{x^3}
$$ ($c_1$ is the integral constant, for large $x$ this term can be neglected).
My question: Is the asymptotic formula correct? If not, then how to find the asymptotic of the function $\Lambda(x)$?
Is the method correct?
Edit: Though the answer comes wrong with the relation \eqref{2} , but if we use the identity involving the equation $A(x)$ instead of ${\zeta_x}^2(1)=\tau(x)$, then we get the correct answer (the leading term). The approximation \eqref{2} works well here. See my answer   below.
 A: We have, for $x\geq 2$,
\begin{align*}
\sum_{\substack{ 1 \leq m,n \leq x \\ \mathrm{gcd}(m,n)=1}} \frac{1}{mn}
&=\sum_{1 \leq m,n \leq x}\frac{1}{mn}\sum_{k\mid\mathrm{gcd}(m,n)}\mu(k)\\
&=\sum_{1\leq k\leq x}\mu(k)\sum_{\substack{ 1 \leq m,n \leq x \\ k\mid\mathrm{gcd}(m,n)}} \frac{1}{mn}\\
&=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\sum_{1\leq m\leq x/k}\frac{1}{m}\right)^2\\
&=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\log\frac{x}{k}+\gamma+O\left(\frac{k}{x}\right)\right)^2\\
&=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\log^2\frac{x}{k}+2\gamma\log\frac{x}{k}+O(1)\right)\\
&=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\log^2 x-2\log x\log k+2\gamma\log x+O(\log^2 k)\right)\\
&=S_1(x)(\log^2 x+2\gamma\log x)-S_2(x)(2\log x)+O(1),
\end{align*}
where
\begin{align*}
S_1(x)&:=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}=\frac{6}{\pi^2}+O\left(\frac{1}{x}\right),\\
S_2(x)&:=\sum_{1\leq k\leq x}\frac{\mu(k)\log k}{k^2}=\frac{36\zeta'(2)}{\pi^4}+O\left(\frac{\log x}{x}\right).
\end{align*}
We conclude that, for $x\geq 2$,
$$\sum_{\substack{ 1 \leq m,n \leq x \\ \mathrm{gcd}(m,n)=1}} \frac{1}{mn}=
\frac{6}{\pi^2}\log^2 x+C\log x+O(1),$$
where
$$C:=\frac{12\gamma}{\pi^2}-\frac{72\zeta'(2)}{\pi^4}=1.3947995\dots$$
