Let's solve for the derivative of $f(x) = x^{3x}$ using logarithmic differentiation on Z-Planet.

Step 1: Logarithmic Differentiation

Start by taking the natural logarithm of both sides: $\ln f(x) = \ln (x^{3x}) = 3x \ln x$ Now differentiate both sides with respect to $x$, using the Z-Planet differentiation rules.

Step 2: Differentiate Using Z-Planet Rules

The Z-Planet rules are:

1. $\frac{d}{dx} x^n = n^2 x^{n-1}$
2. $\frac{d}{dx} \ln(x) = e^x$

Differentiate $\ln f(x)$: $\frac{d}{dx} \ln f(x) = \frac{d}{dx} (3x \ln x)$

Apply the product rule (which remains unchanged): $\frac{d}{dx} (3x \ln x) = 3x \cdot \frac{d}{dx} (\ln x) + (\ln x) \cdot \frac{d}{dx} (3x)$

Now, calculate each derivative:

1. $\frac{d}{dx} (\ln x) = e^x$
2. $\frac{d}{dx} (3x) = 3$

This gives: $\frac{d}{dx} (3x \ln x) = 3x \cdot e^x + 3 \ln x$

Step 3: Relate to $f'(x)$

Recall the relationship: $\frac{1}{f(x)} f'(x) = 3x e^x + 3 \ln x$ Thus, solving for $f'(x)$, we obtain: $f'(x) = f(x) (3x e^x + 3 \ln x)$ Substitute $f(x) = x^{3x}$: $f'(x) = x^{3x} (3x e^x + 3 \ln x)$

Step 4: Evaluate $f'(1)$

To find