Let y=tanAtanB,
So A,B>0 and A+B=π3⇒A,B<π3
⇒tanAtanB<3⇒y∈(0,3)⋯(i)
Now
A+B=π3⇒tan(A+B)=√3
⇒tanA+tanB1−tanAtanB=√3
⇒tanA+ytanA1−y=√3
⇒tan2A+√3(y−1)tanA+y=0
For real +ive value of tanA,
3(y−1)2−4y≥0
⇒3y2−10y+3≥0
⇒(y−3)(y−13)≥0
y∈(−∞,13]∪[3,∞)⋯(ii)
∴ from (i) and (ii) y∈(0,13]
So, maximum value of 3y is 1.