$\text{The total number of real solutions of the equation}$
$\theta ={\tan }^{-1}\left(2\tan \theta \right)-\frac{1}{2}{\sin }^{-1}\left(\frac{6\tan \theta }{9+{\tan }^2\theta }\right)\text{is}$

$\text{(Here, the inverse trigonometric functions}{\sin }^{-1}\text{x and}{\tan }^{-1}\text{x assume values in}[-\frac{\pi }{2},\frac{\pi }{2}]\text{and}(-\frac{\pi }{2},\frac{\pi }{2})\text{, respectively.)}$

The given equation is:
$\theta ={\tan }^{-1}\left(2\tan \theta \right)-\frac{1}{2}{\sin }^{-1}\left(\frac{6\tan \theta }{9+{\tan }^2\theta }\right)$

Let $t=\tan \theta$. We can rewrite the argument of the ${\sin }^{-1}$ term as:
$\frac{6t}{9+t^2}=\frac{2\left({\displaystyle \frac{t}{3}}\right)}{1+{\left({\displaystyle \frac{t}{3}}\right)}^2}$

Let $x=\frac{t}{3}=\frac{\tan \theta }{3}$. The standard definition for the inverse sine identity is:
${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=\{\begin{array}{l}2{\tan }^{-1}x,\text{if}\left|x\right|\le 1\\ \pi -2{\tan }^{-1}x,\text{if}x>1\\ -\pi -2{\tan }^{-1}x,\text{if}x<-1\end{array}$

Let us analyze the equation by dividing it into different intervals based on the value of $x$:

Case 1: $\left|x\right|\le 1$ (which means $\left|\tan \theta \right|\le 3$)
In this interval:
$\frac{1}{2}{\sin }^{-1}\left(\frac{6\tan \theta }{9+{\tan }^2\theta }\right)={\tan }^{-1}\left(\frac{\tan \theta }{3}\right)$
Substituting this back into the original equation, we get:
$\theta ={\tan }^{-1}\left(2\tan \theta \right)-{\tan }^{-1}\left(\frac{\tan \theta }{3}\right)$

Taking the tangent function on both sides:
$\tan \theta =\tan \left[{\tan }^{-1}\right(2\tan \theta )-{\tan }^{-1}(\frac{\tan \theta }{3}\left)\right]$
Letting $t=\tan \theta$:
$t=\frac{2t-{\displaystyle \frac{t}{3}}}{1+2t\left({\displaystyle \frac{t}{3}}\right)}$
$t=\frac{5t}{3+2t^2}$
$t(3+2t^2)=5t$
$2t^3-2t=0$
$2t(t^2-1)=0$
This yields the roots:
$t=0,t=1,t=-1$

Since $\left|t\right|\le 3$ holds true for all these values, we determine the corresponding values of $\theta$ in the principal domain:
1. If $t=0\Rightarrow \tan \theta =0\Rightarrow \theta =0$.
2. If $t=1\Rightarrow \tan \theta =1\Rightarrow \theta =\frac{\pi }{4}$.
3. If $t=-1\Rightarrow \tan \theta =-1\Rightarrow \theta =-\frac{\pi }{4}$.
All three are valid real solutions.

Case 2: $x>1$ (which means $\tan \theta >3$)
In this interval:
$\frac{1}{2}{\sin }^{-1}\left(\frac{6\tan \theta }{9+{\tan }^2\theta }\right)=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{\tan \theta }{3}\right)$
The equation becomes:
$\theta ={\tan }^{-1}\left(2\tan \theta \right)-\frac{\pi }{2}+{\tan }^{-1}\left(\frac{\tan \theta }{3}\right)$
Rearranging:
$\theta +\frac{\pi }{2}={\tan }^{-1}\left(2\tan \theta \right)+{\tan }^{-1}\left(\frac{\tan \theta }{3}\right)$
Taking the tangent on both sides:
$-\cot \theta =\frac{2t+{\displaystyle \frac{t}{3}}}{1-2t\left({\displaystyle \frac{t}{3}}\right)}$
$-\frac{1}{t}=\frac{7t}{3-2t^2}$
$-3+2t^2=7t^2\Rightarrow 5t^2=-3$
Since $t^2\ge 0$ for real numbers, there are no real solutions in this case.

Case 3: $x<-1$ (which means $\tan \theta <-3$)
By symmetry, this case also leads to the equation $5t^2=-3$, yielding no real solutions.

Thus, the only real solutions to the equation are $\theta =-\frac{\pi }{4},0,\frac{\pi }{4}$. The total number of solutions is 3.