Find the derivative of the function.
y = tan(9θ)^2
y'(θ) = ?
$\displaystyle y = \tan^2{\theta} = (\tan{\theta})^2$.
Let $\displaystyle u = \tan{\theta}$, then $\displaystyle y = u^2$.
Now use the chain rule, $\displaystyle \frac{dy}{d\theta} = \frac{du}{d\theta}\times \frac{dy}{du}$.
$\displaystyle \frac{du}{d\theta} = \sec^2{\theta}$
$\displaystyle \frac{dy}{du} = 2u = 2\tan{\theta}$
So $\displaystyle \frac{dy}{d\theta} = 2\tan{\theta}\sec^2{\theta}$.
Hope that helped.