Answer
$\displaystyle \frac{|(x-3)^{1/3}|}{3(x-3)}$
Work Step by Step
Apply the Generalized Rule for absolute values,
$\displaystyle \frac{d}{dx}|w|=\frac{|w|}{w}\frac{dw}{dx}=\frac{|(x-3)^{1/3}|}{(x-3)^{1/3}}\cdot\frac{dw}{dx}$
$w(x)=(x-3)^{1/3}$
$w$ is a composite function $u(v(x))$ where
$u(x)=x^{1/3},\quad v(x)=x-3$
$\displaystyle \frac{du}{dv}=\frac{1}{3}v^{-2/3},\qquad \frac{dv}{dx}=1$
$\displaystyle \frac{dw}{dx}=\frac{du}{dv}\frac{dv}{dx}=\frac{1}{3}v^{-2/3}\cdot 1$
$=\displaystyle \frac{1}{3}(x-3)^{-2/3}$
$\displaystyle \frac{d}{dx}|w|=\frac{|(x-3)^{1/3}|}{(x-3)^{1/3}}\cdot\frac{1}{3}(x-3)^{-2/3}$
$=\displaystyle \frac{|(x-3)^{1/3}|}{3(x-3)^{1/3+2/3}}$
$=\displaystyle \frac{|(x-3)^{1/3}|}{3(x-3)}$