The integral \( \int_{0}^{\pi} \tan^2 \left( \frac{\theta}{3} \right) d\theta \) is to be evaluated. Using the identity \( \tan^2 \theta = \sec^2 \theta - 1 \), the integral transforms to:
\[
\int_{0}^{\pi} \left( \sec^2 \left( \frac{\theta}{3} \right) - 1 \right) d\theta
\]
This can be separated into two integrals:
\[
\int_{0}^{\pi} \sec^2 \left( \frac{\theta}{3} \right) d\theta - \int_{0}^{\pi} 1 \, d\theta
\]
For the first integral, let \( u = \frac{\theta}{3} \), so \( du = \frac{1}{3} d\theta \). The limits of integration change from \( \theta = 0 \) to \( u = 0 \) and from \( \theta = \pi \) to \( u = \frac{\pi}{3} \). The first integral becomes:
\[
\int_{0}^{\pi/3} 3 \sec^2 u \, du = 3 \left[ \tan u \right]_{0}^{\pi/3} = 3 \left( \tan \left( \frac{\pi}{3} \right) - \tan(0) \right) = 3 \times \sqrt{3}
\]
The second integral is:
\[
\int_{0}^{\pi} 1 \, d\theta = [\theta]_{0}^{\pi} = \pi
\]
Combining the results of the two integrals yields:
\[
3\sqrt{3} - \pi
\]
Therefore, the value of the integral is \( 3\sqrt{3} - \pi \).