Ignoring constants when finding derivatives of trig functions

If you'd like to use the product rule, then use the product rule. Let's do it now.

The $(fg)' = f'g + fg'$, so $$\frac{d}{dx} 3\sin^2(6x) = \left(\frac{d}{dx} 3\right) \sin^2(6x) + 3\left(2\sin(6x)\cos(6x)6\right) = 0 + 3\left(12\sin(6x)\cos(6x)\right).$$

You'll notice that it's just as if the $3$ factored out. While I'd recommend recognizing that you can pull constants out, it is entirely possible to use the product rule every time.

You also ask: How come you wouldn't always pull the constant out? My answer - I always would pull the constant out. Let's even prove that this happens:

A derivative of a function $f$ is

$$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h},$$

so that

$$ \frac{d}{dx} cf(x) = \lim_{h \to 0} \frac{cf(x+h) - cf(x)}{h} = c\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = cf'(x).$$

More generally, differentiation is a linear operator, meaning that $(af(x) + bg(x))' = af'(x) + bg'(x)$ for any functions $f,g$ and constants $a,b$.