Scaling property of Fourier Transform
Ok you are mixing the order of the transforms. When you decompose the argument $3t-6$ into $3\cdot(t-2)$ , then you should first scale $x(t)$ by $3$ and then shift the scaled result by $2$, that's where you have chosen the wrong order. Let's show the correct order:

Let $x(t)$ be the original signal whose CTFT is $X(\omega)$, then define the following signals in two steps to get the final signal $z(t)= x(3t-6)$:

$$ y(t) = x(3t) \tag{1}$$ and
$$z(t) = y(t-2) \implies z(t) =  x(3(t-2)) = x(3t-6) \tag{2}$$

Notice how we replaced $y(t-2)$ by using its definition from the eq. (1).

Now applying CTFT to the above rows yields:

$$ Y(\omega) = \frac{1}{3} X(\omega/3) \tag{3} $$ and
$$ Z(\omega) = Y(\omega) e^{-j2\omega} \tag{4} $$

finally insert $Y(\omega)$ from eq. (3) into eq. (4) to get:

$$ \boxed{Z(\omega) = Y(\omega) e^{-j2\omega} = \frac{1}{3} X(\omega/3) e^{-j2\omega} }$$

By the way, you could have also used the alternate approach: first shift $x(t)$ by $6$: $$y(t) = x(t-6) \implies Y(\omega) = X(\omega)e^{-j6\omega}$$ and then scale $y(t)$ by $3$ to obtain $$z(t) = y(3t) = x(3t-6) \implies Z(\omega) = \frac{1}{3}Y(\omega/3) = \frac{1}{3}X(\omega/3)e^{-j2\omega}$$