To solve this limit using L'Hôpital's Rule, we need to check that it fits the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, where L'Hôpital's Rule applies. Let's proceed with this step-by-step:

$\lim_{{x \to 3}} \left( \frac{4}{\ln(x + 4)} - \frac{4x}{x + 3} \right)$

Step 1: Rewrite as a Single Fraction

To apply L'Hôpital's Rule effectively, we should combine the two terms into a single fraction:

$\lim_{{x \to 3}} \left( \frac{4(x + 3) - 4x \ln(x + 4)}{(x + 3) \ln(x + 4)} \right)$

Expanding the numerator:

$= \frac{4x + 12 - 4x \ln(x + 4)}{(x + 3) \ln(x + 4)}$

Now, substituting $x = 3$ directly, we get $0/0$, which means L'Hôpital's Rule is applicable.

Step 2: Differentiate the Numerator and Denominator

Using L'Hôpital's Rule, we differentiate the numerator and the denominator with respect to $x$.

1. Numerator: Differentiate $4(x + 3) - 4x \ln(x + 4)$ with respect to $x$: $\frac{d}{dx} [4(x + 3) - 4x \ln(x + 4)] = 4 - 4 \ln(x + 4) - \frac{4x}{x + 4}$

2. Denominator: Differentiate $(x + 3) \ln(x + 4)$ with respect to $x$: [