Answer
$f'\left( z \right) = 1 + {e^z} + z - z{e^z}$
Work Step by Step
$$\eqalign{
& f\left( z \right) = \left( {1 - {e^z}} \right)\left( {z + {e^z}} \right) \cr
& {\text{Differentiating}} \cr
& f'\left( z \right) = \frac{d}{{dz}}\left[ {\left( {1 - {e^z}} \right)\left( {z + {e^z}} \right)} \right] \cr
& {\text{Differentiate by using the formula }}\left( {gh} \right)' = gh' + hg' \cr
& {\text{Let }}g = \left( {1 - {e^z}} \right){\text{ and }}h = \left( {z + {e^z}} \right),{\text{ then}} \cr
& f'\left( z \right) = \left( {1 - {e^z}} \right)\left( {z + {e^z}} \right)' + \left( {z + {e^z}} \right)\left( {1 - {e^z}} \right)' \cr
& {\text{Compute derivatives}} \cr
& f'\left( z \right) = \left( {1 - {e^z}} \right)\left( {1 + {e^z}} \right) + \left( {z + {e^z}} \right)\left( {1 - {e^z}} \right) \cr
& {\text{Multiply and simplify}} \cr
& f'\left( z \right) = 1 + {e^z} - {e^z} + {e^{2z}} + z + {e^z} - z{e^z} - {e^{2z}} \cr
& f'\left( z \right) = 1 + {e^z} + z - z{e^z} \cr} $$
