If $\tan^{-1}(-3x)+\tan^{-1}(-2x)=\frac{π}{4}$, then the values of $x$ are

$\frac{-1}{6}$

$\tan^{-1}(-3x)+\tan^{-1}(-2x)=\frac{π}{4}$

$=\tan^{-1}\left(\frac{-3x-2x}{1-(-3x)(-2x)}\right)=\frac{π}{4}$   $\left[∵\tan^{-1}(a)+\tan^{-1}(b)=\tan^(\frac{a+b}{ 1- ab})\right] a>0,b>0 , a.b<1$

applying $\tan$ on both side

we get $\frac{-5x}{1-6x^2}=1$  $[∵\tan\frac{π}{4}=1]$

so $-5x=1-6x^2$

$⇒6x^2-5x-1=0$

$⇒6x^2-6x+x-1=0$

$⇒6x(x-1)+1(x-1)=0$

$⇒(6x+1)(x-1)=0$

$⇒x=1,\frac{-1}{6}$

$\text{ only } x= \frac{-1}{6} \text{ Satisfy the condition}$