Ans: $f'(x) = \dfrac{-6x}{(3x^2+4)^2}$
$\begin{array}{cl}
& f'(x) \\
= & \dlim_{h\to 0} \dfrac{f(x+h)-f(h)}{h} \\
= & \dlim_{h \to 0} \dfrac{1}{h}\left( \dfrac{1}{3(x+h)^2+4} -\dfrac{1}{3x^2+4}\right) \\
= & \dlim_{h \to 0} \dfrac{1}{h}\left( \dfrac{3x^2+4-3(x+h)^2-4}{[3(x+h)^2+4](3x^2+4)} \right) \\
= & \dlim_{h \to 0} \dfrac{1}{h}\left( \dfrac{3x^2+4-3x^2-6xh-3h^2-4}{[3(x+h)^2+4](3x^2+4)} \right) \\
= & \dlim_{h \to 0} \dfrac{1}{h}\left( \dfrac{-6xh-3h^2}{[3(x+h)^2+4](3x^2+4)} \right) \\
= & \dlim_{h \to 0} \dfrac{-6x-3h}{[3(x+h)^2+4](3x^2+4)} \\
= & \dfrac{-6x-3(0)}{[3(x+0)^2+4](3x^2+4)} \\
= & \dfrac{-6x}{(3x^2+4)^2}
\end{array}$
