To solve this problem, we need to differentiate the given integral with respect to $x$. The problem is expressed as:

$\frac{d}{dx} \int_{-3x^2}^{\ln(3x)} e^{2t} \, dt$

This expression involves the differentiation of a definite integral where the limits of integration depend on $x$. We will use Leibniz's rule, which states:

$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$

Here's how we proceed:

1. Identify the limits: The lower limit is $a(x) = -3x^2$ and the upper limit is $b(x) = \ln(3x)$.
2. Differentiate the limits:
   • $a'(x) = \frac{d}{dx}(-3x^2) = -6x$
   • $b'(x) = \frac{d}{dx}(\ln(3x)) = \frac{1}{x}$
3. Evaluate the integrand $e^{2t}$ at the limits:
   • At the upper limit $t = b(x) = \ln(3x)$, we have $f(b(x)) = e^{2\ln(3x)} = (3x)^2 = 9x^2$.
   • At the lower limit $t = a(x) = -3x^2$, we have $f(a(x)) = e^{2(-3x^2)} = e^{-6x^2}$.

Applying the formula:

$\frac{d}{dx} \int_{-3x^2}^{\ln(3x)} e^{2t} \, dt = (3x)^2 \cdot \frac{1}{x} - e^{-6x^2} \cdot (-6x)$

Simplifying this:

$\frac{d}{dx} \int_{-3x^2}^{\ln(3x)} e^{2t} \, dt = 9x - (-6x) \cdot e^{-6x^2}$

$= 9x + 6x e^{-6x^2}$

Thus, the final answer is:

$\boxed{9x + 6x e^{-6x^2}}$

Would you like a more detailed step-by-step explanation? Here are some related questions to consider:

1. How does Leibniz's rule work when differentiating an integral with variable limits?
2. What are some common mistakes when applying Leibniz's rule?
3. How does differentiation interact with exponential functions?
4. How would the solution change if the integrand were different, such as a trigonometric function?
5. How can you graph the resulting function to understand its behavior?

Tip: Always carefully identify and differentiate the limits when applying Leibniz's rule to avoid sign errors!