$2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)$ The problem


Given that $a,b>0$ and $$2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)$$
Find the value of $$\log _{a b}\left(\frac{1}{a}+\frac{1}{b}\right)$$

My attempt

We have from the given condition
$$2+\frac{\log a}{\log 2}=3+\frac{\log b}{\log 3}=\frac{\log (a+b)}{\log 6}$$
$\implies$
$$\frac{2\log 2+\log a}{\log 2}=\frac{3 \log 3+\log b}{\log 3}=\frac{\log (a+b)}{\log 6}$$
By ratio and proportion we get each ratio equal to
$$\frac{2\log 2+3\log 3+\log(ab)-\log(a+b)}{0}$$
Thus we have
$$\log\left(\frac{1}{a}+\frac{1}{b}\right)=\log(108)$$
But i am unable to find $\log(ab)$

Set
\begin{align}\\ \frac{2\log 2+\log a}{\log 2}=\frac{3 \log 3+\log b}{\log 3}=\frac{\log (a+b)}{\log 6}=y\\\to \log\left(\frac{1}{a}+\frac{1}{b}\right)=\log(108) \to 108ab=a+b\end{align}
This is where you are.
Let us find another equation to  eliminate  a or b.
From  \begin{align}\\2\log 2\log 3+\log a\log 3=3 \log 3\log 2+\log b\log 2\end{align} \begin{align}a=2b^{\frac{\ln 2}{\ln 3}}, 108\cdot 2b^{\frac{\ln 2+\ln 3}{\ln 3}}=2b^{\frac{\ln 2}{\ln 3}}+b\end{align}
Therefore
\begin{align}b \approx 0.01, a\approx 0.11, a+b \approx 0.12, ab\approx 0.0011
\\log_{ab}(\frac{1}{a}+\frac{1}{b})=\log_{ab}{(a+b)} -1\approx -0.6883\end{align}
