Answer
The amplitude and period are the same but for shift, $g$ is shifted $\frac{\pi}{3}$ to the right from $f$.
Work Step by Step
The standard form to use is:
$$y=a\sin b(x-c)$$
For function $f(x)=\sin 3x=1\sin 3(x-0)$:
$a=1$
$period=\frac{2\pi}{b}=\frac{2\pi}{3}$
$shift=none$
For function $g(x)=\sin (-3x)=\sin (3x-\pi)=1\sin 3\left(x-\frac{\pi}{3}\right)$:
$a=1$
$period=\frac{2\pi}{b}=\frac{2\pi}{3}$
$shift=\frac{\pi}{3}~to~the ~right$
Thus, the amplitude and period are the same but for shift, $g$ is shifted $\frac{\pi}{3}$ to the right from $f$.
