Rewrite f(x) as a nested exponential-logarithm expression :
[tex]\left(\ln(3x)\right)^{2x} = \exp\left(\ln\left(\ln(3x)\right)^{2x}\right)[/tex]
(where [tex]\exp(x) = e^x[/tex])
One of the properties of logarithms lets us drop the exponent as a coefficient:
[tex]\exp\left(\ln\left(\ln(3x)\right)^{2x}\right) = \exp\left(2x\ln\left(\ln(3x)\right)\right)[/tex]
Now, by the chain rule, we have
[tex]f(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \implies \\\\ f'(x) = \left(\exp\left(2x\ln\left(\ln(3x)\right)\right)\right)' \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2x\ln\left(\ln(3x)\right)\right)'[/tex]
By the product rule,
[tex]f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left( (2x)' \ln\left(\ln(3x)\right) + 2x\left(\ln\left(\ln(3x)\right)\right)' \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left( 2\ln\left(\ln(3x)\right) + 2x\left(\ln\left(\ln(3x)\right)\right)' \right)[/tex]
By the chain rule again,
[tex]f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + 2x \cdot \dfrac1{\ln(3x)} \cdot \left(\ln(3x)\right)' \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2x}{\ln(3x)} \cdot \dfrac1{3x} \cdot (3x)' \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{3\ln(3x)} \cdot 3 \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{\ln(3x)} \right)[/tex]
Then simplify this to
[tex]f'(x) = \boxed{\left(\ln(3x)\right)^{2x} \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{\ln(3x)} \right)}[/tex]
