$\text{The total number of real solutions of the equation}$
$\theta ={\tan }^{-1}\left(2\tan \theta \right)-\frac{1}{2}{\sin }^{-1}\left(\frac{6\tan \theta }{9+{\tan }^2\theta }\right)\text{is}$
$\text{(Here, the inverse trigonometric functions}{\mathrm{s\; in}}^{-1}\text{x and}{\mathrm{ tan}}^{-1}\text{x assume values in}[-\frac{\pi }{2},\frac{\pi }{2}]\text{and}(-\frac{\pi }{2},\frac{\pi }{2})\text{, respectively.)}$

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

Let $x=\tan \theta$. First, let us simplify the term inside the inverse sine function:
$\frac{6\tan \theta }{9+{\tan }^2\theta }=\frac{6x}{9+x^2}=\frac{2(x/3)}{1+{(x/3)}^2}$

We use the standard property of the inverse sine function:
${\sin }^{-1}\left(\frac{2u}{1+u^2}\right)=2{\tan }^{-1}\left(u\right)\text{ for }\left|u\right|\le 1$
Here, $u=\frac{x}{3}$. Let us check the case where $\left|u\right|\le 1$, which corresponds to $\left|\tan \theta \right|\le 3$.

Under this condition, the term simplifies to:
$\frac{1}{2}{\sin }^{-1}\left(\frac{6x}{9+x^2}\right)=\frac{1}{2}·2{\tan }^{-1}\left(\frac{x}{3}\right)={\tan }^{-1}\left(\frac{x}{3}\right)$

Substituting this back into the original equation, we get:
$\theta ={\tan }^{-1}\left(2x\right)-{\tan }^{-1}\left(\frac{x}{3}\right)$

Since the range of ${\tan }^{-1}\left(2x\right)$ is $(-\frac{\pi }{2},\frac{\pi }{2})$ and the range of ${\tan }^{-1}\left(\frac{x}{3}\right)$ is also $(-\frac{\pi }{2},\frac{\pi }{2})$, the RHS lies within $(-\pi ,\pi )$. For any solution, we must have $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$ for the tangent function on the LHS to be defined. Taking the tangent of both sides:
$\tan \theta =\tan \left({\tan }^{-1}\left(2x\right)-{\tan }^{-1}\left(\frac{x}{3}\right)\right)$
$x=\frac{2x-\frac{x}{3}}{1+2x·\frac{x}{3}}$
$x=\frac{\frac{5x}{3}}{1+\frac{2x^2}{\mathrm{3<!--\; <mrow-->}}}$
$x=\frac{5x}{3+2x^2}$

Rearranging this equation:
$x(3+2x^2)=5x$
$3x+2x^3-5x=0$
$2x^3-2x=0$
$2x(x^2-1)=0$

This yields three real values for $x$:
$x=0,x=1,\text{ and }x=-1$

All three values satisfy our initial assumption $\left|x\right|\le 3$. Let us find the corresponding values of $\theta$ in the principal domain $(-\frac{\pi }{2},\frac{\pi }{2})$:
1. For $x=0\Rightarrow \tan \theta =0\Rightarrow \theta =0$.
2. For $x=1\Rightarrow \tan \theta =1\Rightarrow \theta =\frac{\pi }{4}$.
3. For $x=-1\Rightarrow \tan \theta =-1\Rightarrow \theta =-\frac{\pi }{4}$.

Analyzing the alternative intervals (e.g. $\left|u\right|>1$) yields no further valid real solutions. Therefore, the total number of real solutions of the equation is 3.