1. **State the problem:** Find the derivative $f'(x)$ of the function $$f(x) = 2x^3 \ln^2 x.$$\n\n2. **Recall the formula and rules:** We will use the product rule for derivatives since $f(x)$ is a product of two functions: $u(x) = 2x^3$ and $v(x) = \ln^2 x$. The product rule states: $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).$$\nAlso, recall the chain rule for $v(x) = (\ln x)^2$: $$v'(x) = 2\ln x \cdot \frac{1}{x} = \frac{2\ln x}{x}.$$\n\n3. **Calculate derivatives of each part:**\n- Derivative of $u(x) = 2x^3$ is $$u'(x) = 6x^2.$$\n- Derivative of $v(x) = \ln^2 x$ is $$v'(x) = \frac{2\ln x}{x}.$$\n\n4. **Apply the product rule:**\n$$f'(x) = u'(x)v(x) + u(x)v'(x) = 6x^2 \ln^2 x + 2x^3 \cdot \frac{2\ln x}{x}.$$\n\n5. **Simplify the second term:**\n$$2x^3 \cdot \frac{2\ln x}{x} = 2x^3 \cdot \frac{2\ln x}{x} = 4x^2 \ln x.$$\n\n6. **Write the final derivative:**\n$$f'(x) = 6x^2 \ln^2 x + 4x^2 \ln x = 2x^2 (3 \ln^2 x + 2 \ln x).$$\n\n**Answer:** $$\boxed{f'(x) = 6x^2 \ln^2 x + 4x^2 \ln x}.$$