This is the question,
When I am solving this I am getting the equation
$$(2x)(3x+2)(x-3)=0$$
When I am putting $x=-\frac{2}{3}$ and $x=3$ the equation is satisfied and when I am putting $x=0$ the equation goes like $$\frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$$
Clearly, it satisfies the equation but can I do it this way, as the range of tan inverse x is ($\frac{\pi}{2}, \frac{\pi}{2}$).
Moreover, can you check if the $x=-2/3$ satisfies the equation?
This is how I am solving it,
When $x=-2/3$, then LHS$<0$ and RHS$>0$.
With $A=\arctan \frac{1}{2x+1}$, $B=\arctan \frac{1}{4x+1}$, $C=\arctan \frac{2}{x^2}$, what you solve is $$\tan (A+B)=\tan C,$$ and it may happen that $A+B=C+k\pi$ for some $k\in\mathbb Z$.
Just check each possible answer.
The solution $3$ is OK, since $A>0$, $B>0$, $C>0$ and $A+B< \pi$.
With $x=-\frac{2}{3}$, $A,B<0$, $C>0$, so we must have $A+B=C-\pi$.