Step 1: Read what the equation is asking.
We are told that $\tan^{-1}(3x)+\tan^{-1}(2x)=\frac{\pi}{4}$. Rather than jumping into a formula, let us name the two angles so the idea stays clear. Let $A=\tan^{-1}(3x)$ and $B=\tan^{-1}(2x)$, so that $\tan A=3x$ and $\tan B=2x$.
Step 2: Recall what the sum of the angles must be.
Since $A+B=\frac{\pi}{4}$, the tangent of the sum must equal $\tan\frac{\pi}{4}=1$. So our job becomes finding $x$ for which $\tan(A+B)=1$.
Step 3: Expand the tangent of a sum.
Using $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ and substituting the values from Step 1, \[ \frac{3x+2x}{1-(3x)(2x)}=1. \]
Step 4: Clear the fraction.
This gives $\dfrac{5x}{1-6x^2}=1$, so $5x=1-6x^2$. Bringing every term to one side, \[ 6x^2+5x-1=0. \]
Step 5: Solve the quadratic by splitting the middle term.
Write $5x=6x-x$, giving $6x^2+6x-x-1=0$, which factors as $6x(x+1)-1(x+1)=0$, that is $(6x-1)(x+1)=0$. So $x=\frac16$ or $x=-1$.
Step 6: Keep only the value that fits the original equation.
If $x=-1$ then both $\tan^{-1}(-3)$ and $\tan^{-1}(-2)$ are negative, so their sum is negative and cannot equal the positive angle $\frac{\pi}{4}$. Hence we reject $x=-1$ and keep $x=\frac16$.
\[ \boxed{x=\frac16} \]