1. Problem: Find the derivative $\frac{dy}{dx}$ of the function $y = x^x \cosh^2(x^3)$.\n\n2. Formula and rules: \n- Use the product rule: $\frac{d}{dx}[u v] = u' v + u v'$.\n- Use the chain rule for composite functions.\n- Derivative of $x^x$ is $x^x (\ln x + 1)$.\n- Derivative of $\cosh^2(u)$ is $2 \cosh(u) \sinh(u) \frac{du}{dx}$.\n\n3. Let $u = x^x$ and $v = \cosh^2(x^3)$.\n- Compute $u' = \frac{d}{dx} x^x = x^x (\ln x + 1)$.\n- Compute $v' = \frac{d}{dx} \cosh^2(x^3)$.\n - Let $w = x^3$, then $v = \cosh^2(w)$.\n - $\frac{dv}{dw} = 2 \cosh(w) \sinh(w)$ and $\frac{dw}{dx} = 3x^2$.\n - So, $v' = 2 \cosh(x^3) \sinh(x^3) \cdot 3x^2 = 6x^2 \cosh(x^3) \sinh(x^3)$.\n\n4. Apply product rule:\n$$\frac{dy}{dx} = u' v + u v' = x^x (\ln x + 1) \cosh^2(x^3) + x^x \cdot 6x^2 \cosh(x^3) \sinh(x^3)$$\n\n5. Final answer:\n$$\boxed{\frac{dy}{dx} = x^x (\ln x + 1) \cosh^2(x^3) + 6x^{x+2} \cosh(x^3) \sinh(x^3)}$$