If $x>0, y>0,x+y=\frac{\pi}{3}$ then maximum value of $\tan x\tan y$ 
If $x>0, y>0$ and $x+y=\frac{\pi}{3}$, then find the maximum value of $\tan x\tan y$

My Attempt
$x>0, y>0, x+y=\frac{\pi}{3}\implies x, y$ in $1^\text{st}$ quadrant.
$\tan x, \tan y>0, \tan(x+y)=\sqrt{3}, \tan x\tan y>0$
$$
\tan x+\tan y\geq2\sqrt{\tan x.\tan y}\implies \tan^2x+\tan^2y+2\tan x\tan y\geq 4\tan x\tan y\\
2\tan x\tan y\leq\tan^2x+\tan^2y\implies \color{red}{?}
$$
or
$$
1-\tan x\tan y=\frac{\tan x+\tan y}{\tan(x+y)}\implies\tan x\tan y=1-\frac{\tan x+\frac{\sqrt{3}-\tan x}{1+\sqrt{3}\tan x}}{\sqrt{3}}\\
\tan x\tan y=1-\frac{\frac{\tan x+\sqrt{3}\tan^2x+\sqrt{3}-\tan x}{1+\sqrt{3}\tan x}}{\sqrt{3}}=1-\frac{1+\tan^2x}{1+\sqrt{3}\tan x}\leq\color{red}{?}\\
=\frac{1+\sqrt{3}\tan x-1-\tan^2x}{1+\sqrt{3}\tan x}=\frac{\tan x(\sqrt{3}-\tan x)}{1+\sqrt{3}\tan x}=\color{red}{?}
$$
Note: I prefer not to do differentiation

For $-\dfrac{\pi}{6} < t < \dfrac{\pi}{6}$, let $x = \dfrac{\pi}{6} - t \ $ and $ \ y = \dfrac{\pi}{6} + t$.
Then $x > 0$, $y>0$ and $x + y = \dfrac{\pi}{3}$.
Let
\begin{align}
   f(t) 
   &= \tan(x) \tan(y) \\
   &= \tan\left(\dfrac{\pi}{6} - t \right) \tan \left(\dfrac{\pi}{6} + t \right) \\
   &= \dfrac{1-\sqrt 3 \tan t}{\sqrt 3 + \tan t} \cdot
      \dfrac{1+\sqrt 3 \tan t}{\sqrt 3 - \tan t} \\
   &= \dfrac{1 - 3 \tan^2 t}{3 - \tan^2 t} \\
   &= \dfrac{3 \tan^2 t - 1}{\tan^2 t - 3} \\
   &= 3 - \dfrac{8}{3 - \tan^2 t}
\end{align}
Note that $f(t)$ is an even function on the interval $\left[-\dfrac{\pi}{6}, \dfrac{\pi}{6} \right]$. 
\begin{align}
   t \in \left[-\dfrac{\pi}{6}, 0 \right]
   &\implies -\dfrac{1}{\sqrt 3} \le \tan t \le 0 \\
   &\implies 0 \le \tan^2 t \le \dfrac 13 \\
   &\implies \dfrac 83 \le 3 - \tan^2 t \le 3 \\
   &\implies  \dfrac 83 \le \dfrac{8}{3 - \tan^2 t} \le 3 \\
   &\implies 0 \le f(t) \le \dfrac 13
\end{align}
