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.

`\[CapitalLambda][x_]:=Sum[1/Times@@mn,{mn,Select[Tuples[Range[x],2],CoprimeQ@@#&]}]`

)$$x^3\left(\Lambda(x)-\left(\frac{2}{3}\ln^2(x)+\left(\frac{8}{9}+\frac{4}{3}\gamma\right)\ln(x)+\left(\frac{2}{3}\gamma^2+\frac{8}{9}\gamma-\frac{8}{27}\right)\right)\right)$$goes like$$0.561097,0.72963,-1.77106,-22.5681,-41.879,-140.577,-202.573,-384.69,-575.653,-962.996,-1180.86,-1858.96,-2198.3,-3077.97,-3986.3,-5207.03,-5898.68,-7809.84,-8719.26,-10982.,-13018.1,-15792.3,-17282.4,-21169.4,-23506.3,...$$ Does not look like this will tend to a finite limit $c_1$ $\endgroup$1more comment