#### Answer

$f'(z)=3z^{2}e^{3z}+2z(e^{3z}+4)+\dfrac{2(z^{2}-1)}{(z^{2}+1)^{2}}$

#### Work Step by Step

$f(z)=z^{2}(e^{3z}+4)-\dfrac{2z}{z^{2}+1}$
Evaluate the derivative by applying the product rule on the first term and the quotient rule on the second term:
$f'(z)=z^{2}(e^{3z}+4)'+(e^{3z}+4)(z^{2})'-\dfrac{(z^{2}+1)(2z)'-(2z)(z^{2}+1)'}{(z^{2}+1)^{2}}=...$
Evaluate the indicated derivatives:
$...=3z^{2}e^{3z}+2z(e^{3z}+4)-\dfrac{2(z^{2}+1)-2z(2z)}{(z^{2}+1)^{2}}=...$
Simplify the fraction:
$...=3z^{2}e^{3z}+2z(e^{3z}+4)-\dfrac{2z^{2}+2-4z^{2}}{(z^{2}+1)^{2}}=...$
$...=3z^{2}e^{3z}+2z(e^{3z}+4)-\dfrac{-2z^{2}+2}{(z^{2}+1)^{2}}=...$
$...=3z^{2}e^{3z}+2z(e^{3z}+4)+\dfrac{2(z^{2}-1)}{(z^{2}+1)^{2}}$