#### Answer

$w' = \frac{-(z^{2}+1)}{z^{2}}=-1-\frac{1}{z^2}$
$w'' = \frac{2}{z^{3}}$

#### Work Step by Step

$w = (\frac{1+3z}{3z})(3-z)$
$w = \frac{3+8z-3z^{2}}{3z}$
$w' = \frac{3z(8-6z) - 3(3+8z-3z^{2})}{9z^{2}}$
$w' = \frac{-9(z^{2}+1)}{9z^{2}}$
$w' = \frac{-(z^{2}+1)}{z^{2}}$
$w'' = \frac{-z^{2}(2z)+2z(z^{2}+1)}{z^{4}}$
$w'' = \frac{2z}{z^{4}}$
$w'' = \frac{2}{z^{3}}$