1. **State the problem:** We need to find the derivative $\frac{dy}{dx}$ of the function $$y = (4x^2 + 3)^3 e^{(3x^2 + 2)^2}.$$\n\n2. **Identify the formula and rules:** This is a product of two functions: $$u = (4x^2 + 3)^3$$ and $$v = e^{(3x^2 + 2)^2}.$$ We will use the product rule: $$\frac{d}{dx}(uv) = u'v + uv'.$$\n\n3. **Find $u'$:** Using the chain rule, $$u = (4x^2 + 3)^3,$$ so $$u' = 3(4x^2 + 3)^2 \cdot \frac{d}{dx}(4x^2 + 3) = 3(4x^2 + 3)^2 \cdot 8x = 24x(4x^2 + 3)^2.$$\n\n4. **Find $v'$:** For $$v = e^{(3x^2 + 2)^2},$$ use the chain rule: $$v' = e^{(3x^2 + 2)^2} \cdot \frac{d}{dx}[(3x^2 + 2)^2].$$\nCalculate the inner derivative: $$\frac{d}{dx}[(3x^2 + 2)^2] = 2(3x^2 + 2) \cdot \frac{d}{dx}(3x^2 + 2) = 2(3x^2 + 2) \cdot 6x = 12x(3x^2 + 2).$$\nSo, $$v' = e^{(3x^2 + 2)^2} \cdot 12x(3x^2 + 2).$$\n\n5. **Apply the product rule:**\n$$\frac{dy}{dx} = u'v + uv' = 24x(4x^2 + 3)^2 e^{(3x^2 + 2)^2} + (4x^2 + 3)^3 e^{(3x^2 + 2)^2} \cdot 12x(3x^2 + 2).$$\n\n6. **Factor common terms:** Both terms have $$12x (4x^2 + 3)^2 e^{(3x^2 + 2)^2}$$ in common, so\n$$\frac{dy}{dx} = 12x (4x^2 + 3)^2 e^{(3x^2 + 2)^2} \left[2 + (4x^2 + 3)(3x^2 + 2)\right].$$\n\n7. **Simplify inside the bracket:**\nCalculate $$(4x^2 + 3)(3x^2 + 2) = 12x^4 + 8x^2 + 9x^2 + 6 = 12x^4 + 17x^2 + 6.$$\nSo the bracket becomes $$2 + 12x^4 + 17x^2 + 6 = 12x^4 + 17x^2 + 8.$$\n\n**Final answer:**\n$$\boxed{\frac{dy}{dx} = 12x (4x^2 + 3)^2 e^{(3x^2 + 2)^2} \left(12x^4 + 17x^2 + 8\right)}.$$