Answer
$g'\left( \theta \right) = \left( {{{\tan }^2}\theta + \tan \theta - \theta } \right){e^\theta }$
Work Step by Step
$$\eqalign{
& g\left( \theta \right) = {e^\theta }\left( {\tan \theta - \theta } \right) \cr
& {\text{Differentiating}} \cr
& g'\left( \theta \right) = \frac{d}{{d\theta }}\left[ {{e^\theta }\left( {\tan \theta - \theta } \right)} \right] \cr
& {\text{Use the product rule for derivatives }}\left( {fh} \right)' = fh' + hf' \cr
& {\text{Let }}f = {e^\theta }{\text{ and }}h = \tan \theta - \theta ,{\text{ then}} \cr
& g'\left( \theta \right) = {e^\theta }\frac{d}{{d\theta }}\left[ {\tan \theta - \theta } \right] + \left( {\tan \theta - \theta } \right)\frac{d}{{d\theta }}\left[ {{e^\theta }} \right] \cr
& \cr
& {\text{Use the Derivatives formulas }} \cr
& \frac{d}{{dx}}\left[ {\tan x} \right] = {\sec ^2}x,{\text{ }}\frac{d}{{dx}}\left[ {{e^x}} \right] = {e^x} \cr
& {\text{therefore}} \cr
& g'\left( \theta \right) = {e^\theta }\left( {{{\sec }^2}\theta - 1} \right) + \left( {\tan \theta - \theta } \right){e^\theta } \cr
& {\text{Factoring}} \cr
& g'\left( \theta \right) = \left( {{{\sec }^2}\theta - 1 + \tan \theta - \theta } \right){e^\theta } \cr
& g'\left( \theta \right) =(\tan^2\theta+\tan\theta-\theta)e^{\theta}}
$$