I got the answer for $\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}^{3x+1}$, but only by a mistake - how do I solve correctly?

You are good until here where you take an incorrect derivative of the denominator.

\begin{align} \lim \limits_{x \to \infty} \space -{6(3x-1)^2 \over (3x-2)(3x+4)} \end{align}

You don't even have to apply L'Hopitals rule here, you can just compare the coefficients of the $x^2$ terms. Don't forget the negative.

$$\lim_{x\to\infty} -\frac{6(3x-1)^2}{(3x-2)(3x+4)} = \lim_{x\to\infty} -\frac{\color{red}{54}x^2-36x+6}{\color{red}{9}x^2+6x-8} = -\frac{54}{9} = -6$$

Note that the derivative of the denominator $(3x-2)(3x+4)$ is not $(3)\cdot (3)$, but something else. There is no case of L'Hopital of $\frac{\infty}{\infty\cdot\infty}$, only a case of $\frac{\infty}{\infty}$.

Alternative answer. Write it as:

$$\lim_{x\to\infty} \left(1-\frac{6}{3x+4}\right)^{3x+1}$$

Let $z=\frac{3x+4}{6}$ Then $3x+1=6z-3$ so this is:

$$\lim_{z\to\infty} \left(1-\frac{1}{z}\right)^{6z-3}$$

Now, $\left(1-\frac{1}{z}\right)^z\to e^{-1}$ and $\left(1-\frac{1}{z}\right)^{-3}\to 1$. So the limit is $e^{-6}$.