How do I prove that a function decreases/increases on an interval?

you have to solve the inequality $$f'(x)=4(x-3)^3\geq 0$$ or $$f'(x)=4(x-3)^3\le 0$$ thus your function is increasing if $$3\le x<+\infty$$ or decreasing if $$-\infty<x<3$$

You find the extremum by taking the first derivative, not the second. You will still get $x=3$ in this case. Then to show the function is increasing on $]+3,\infty[$, you just need to show the first derivative is positive on the interval. So take the first derivative and show that.

The big theorem connecting the derivative to monotonicity is:

If $f'$ is positive on an open interval, then $f$ is increasing on that interval (in fact, the closure of that interval). Likewise, if $f'$ is negative on an interval, then $f$ is decreasing on that interval.

In your case, the derivative is $f'(x) = 4(x-3)^3$. This is negative on $-\infty < x < 3$ and positive on $3 < x < \infty$. So $f$ is decreasing on $(-\infty,3]$ and increasing on $[3,\infty)$.