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**Hint:**First expand the given expression in right hand side using the formula for expansion of \[{{\tan }^{-1}}x+{{\tan }^{-1}}y\] and then equating the both left hand side and right hand side we will get the quadratic equation so it has two roots you can solve those roots using factorization method and then get the possible value of x.

**Complete step by step answer:**

Given \[{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+{{\tan }^{-1}}\left( \dfrac{x}{3} \right)=\left( \dfrac{\pi }{4} \right)\]

We know that the formulae for \[{{\tan }^{-1}}x+{{\tan }^{-1}}y\] is given by \[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\]

Now apply the above formula we will get,

\[{{\tan }^{-1}}\left( \dfrac{\left( \dfrac{x}{2} \right)+\left( \dfrac{x}{3} \right)}{1-\left( \dfrac{x}{2} \right)\left( \dfrac{x}{3} \right)} \right)=\left( \dfrac{\pi }{4} \right)\]

\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{5x}{6}}{\dfrac{6-{{x}^{2}}}{6}} \right)=\left( \dfrac{\pi }{4} \right)\]

Now shift the inverse tan function or arctan function to right hand side it becomes tan function on moving from left hand side to right hand side

\[\dfrac{5x}{6-{{x}^{2}}}=1\]

\[\Rightarrow 5x=6-{{x}^{2}}\]

\[\Rightarrow {{x}^{2}}+5x-6=0\]

Now solve the quadratic equations we will get two roots

\[\left( x-1 \right)\left( x+6 \right)=0\]

\[\Rightarrow x=1\left[ 0 < x < \sqrt{6} \right]\]

\[\Rightarrow x=-6\] is not the solution for the given equation because in the question it is given that x belongs only to a certain domain that is \[0 < x < \sqrt{6}\]. hence \[x=-6\] is not in the given domain so it is not a possible value for x.

**So \[x=1\] is the only possible solution for x.**

**Note:**In the question it was given \[0 < x < \sqrt{6}\], this means the value of x lies in between the given interval so that the function gets satisfied. Use the formula for \[{{\tan }^{-1}}x+{{\tan }^{-1}}y\] carefully and calculate to get the final answer.

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