If $A>0,B>0$ and $A+B=\frac{\pi }{3}$ then the maximum value of $\tan A\tan B$ is

$A+B=\frac{\pi }{3}\Rightarrow \tan (A+B)=\sqrt{3}$

$\Rightarrow \frac{\tan A+\mathrm{tanB}}{1-\tan A\tan B}=\sqrt{3}$

$\Rightarrow \frac{\tan A+\frac{y}{\tan A}}{1-y}=\sqrt{3}$

where $y=\tan A\tan B$

$\Rightarrow {\tan }^{2}A+\sqrt{3}(y-1)\tan A+y=0$

for real values $\tan A,3{(y-1)}^2-4y\ge 0$

$\Rightarrow 3y^2-10y+3\ge 0\Rightarrow y\le \frac{1}{3}ory\ge 3$

But, A, B > 0 and $A+B=\frac{\pi }{3}\Rightarrow A,B<\frac{\pi }{3}$

$\Rightarrow \tan A\tan B<3\therefore y\le \frac{1}{3}$

$\therefore$ maximum value of y is $\frac{1}{3}$