Given :
$\tan^{-1}(x-1)+\tan^{-1}(x)+\tan^{-1}(x+1)=\tan^{-1}(3x)$
$\Rightarrow \tan^{-1}(x-1)+\tan^{-1}x=\tan^{-1}(3x)-\tan^{-1}(x+1)$
$\Rightarrow \tan^{-1}\big[\large\frac{(x-1)+x}{1-(x-1)x}\big]$$=\tan^{-1}\big[\large\frac{3x-(x+1)}{1+3x(x+1)}\big]$
$\Rightarrow \large\frac{2x-1}{1-x^2+x}=\frac{2x-1}{1+3x^2+3x}$
$(2x-1)(1+3x^2+3x)=(1-x^2+x)(2x-1)$
$\Rightarrow (2x-1)(4x^2+2x)=0$
$\Rightarrow (2x-1)2x(2x+1)=0$
$\Rightarrow x=0,\pm \large\frac{1}{2}$
Hence (d) is the correct answer.