\begin{align*}
\frac{f\left(h+3\right)-f\left(3\right)}{h} & =\frac{\left[3\left(h+3\right)^2-4\left(h+3\right)+1\right]-\left[3\left(3\right)^2-4\left(3\right)+1\right]}{h} \\

& =\frac{3\left(h^2+6h+9\right)-4h-12+1-16}{h}\\

& =\frac{3h^2+18h+27-4h-12+1-16}{h}\\

& =\frac{3h^2+14h}{h}\\

& =\frac{h\left(3h+14\right)}{h}\\
& =3h+14 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}