Step 1: Introduce $\alpha$.

Let $\alpha =\frac{1}{2}{\sin }^{-1}\left(\frac{6\tan \theta }{9+{\tan }^2\theta }\right)$. The equation becomes $\theta +\alpha ={\tan }^{-1}(2\tan \theta )$.

Step 2: Apply $\tan$ and use the sine identity.

$\tan (\theta +\alpha )=2\tan \theta$ and $\sin 2\alpha =\frac{6\tan \theta }{9+{\tan }^2\theta }=\frac{2\tan \alpha }{1+{\tan }^2\alpha }$.

The second relation yields $3\tan \theta +3\tan \alpha {\tan }^2\theta =9\tan \alpha +\tan \alpha {\tan }^2\theta \cdot \tan \alpha$, factoring to $(\tan \theta -3\tan \alpha )(3-\tan \alpha \tan \theta )=0$.

Step 3: Branch into two cases.

Case 1: $\tan \theta =3\tan \alpha$. Substituting into $\tan (\theta +\alpha )=2\tan \theta$ gives a quadratic in $\tan \theta$ whose solutions reduce to $\tan \theta =0,1,-1$, i.e. $\theta =0,\frac{\pi }{4},-\frac{\pi }{4}$.

Case 2: $\tan \alpha \tan \theta =3$. The resulting equation $\frac{3}{\tan \theta }=-5\tan \theta$ has no real solution since the left and right sides have opposite signs.

Step 4: Count.

Three real solutions in total.

$\therefore$ The correct option is (C).