Given, $A+B=\frac{\pi}{3}$
Let $y=\tan A \tan B$
$y=\tan A \tan \left(\frac{\pi}{3}-A\right)$
Differentiating w.r.t A, we get
$\Rightarrow \frac{d y}{d A}=\sec ^2 A \tan \left(\frac{\pi}{3}-A\right)-\sec ^2\left(\frac{\pi}{3}-A\right) \tan A$
For maxima or minima, $\frac{d y}{d A}=0$
$\begin{aligned}
& \therefore \quad \sec ^2 A \tan \left(\frac{\pi}{3}-A\right)-\sec ^2\left(\frac{\pi}{3}-A\right) \tan A=0 \\
& \Rightarrow \quad \sec ^2 A \tan \left(\frac{\pi}{3}-A\right)=\tan A \sec ^2\left(\frac{\pi}{3}-A\right) \\
& \Rightarrow \quad\left(1+\tan ^2 A\right) \tan \left(\frac{\pi}{3}-A\right) \\
& =\tan A\left[1+\tan ^2 A\left(\frac{\pi}{3}-A\right)\right] \\
& \Rightarrow \quad \tan \left(\frac{\pi}{3}-A\right)+\tan ^2 A \tan \left(\frac{\pi}{3}-A\right) \\
& =\tan A+\tan A \tan ^2\left(\frac{\pi}{3}-A\right) \\
& \Rightarrow \quad\left[\tan \left(\frac{\pi}{3}-A\right)-\tan A\right] \\
& {\left[1-\tan A \tan \left(\frac{\pi}{3}-A\right)\right]=0} \\
& \Rightarrow \quad \tan \left(\frac{\pi}{3}-A\right)=\tan A \\
& \Rightarrow \quad \frac{\neq}{2}-A=A \quad \Rightarrow \quad 2 A=\frac{\neq}{2} \\
& \Rightarrow A=\frac{\pi}{6} \\
& \therefore \quad A=B=\frac{\pi}{6}
\end{aligned}$
Maximum value of $\tan A \tan B=\tan \frac{\not}{2} \tan \frac{\pi}{6}$
$\begin{aligned}
& =\left(\tan \frac{\pi}{6}\right)^2 \\
& =\left(\frac{1}{\sqrt{3}}\right)^2=\frac{1}{3}
\end{aligned}$