Differentiate both sides of the equation (x²y³) + 3x³y² = 6 with respect to x.

For the first term, $\frac{d}{dx}[\sin(x^2y^3)]$:
Apply the chain rule. Let $u = x^2y^3$. Then $\frac{d}{dx}[\sin(u)] = \cos(u) \frac{du}{dx}$.
To find $\frac{du}{dx} = \frac{d}{dx}[x^2y^3]$, apply the product rule:

$$\frac{d}{dx}(x^2y^3) = \frac{d}{dx}(x^2) \cdot y^3 + x^2 \cdot \frac{d}{dx}(y^3)$$ $$= (2x)y^3 + x^2(3y^2 \frac{dy}{dx})$$ $$= 2xy^3 + 3x^2y^2 \frac{dy}{dx}$$

So, the derivative of the first term is:

$$\cos(x^2y^3) (2xy^3 + 3x^2y^2 \frac{dy}{dx}) = 2xy^3 \cos(x^2y^3) + 3x^2y^2 \cos(x^2y^3) \frac{dy}{dx}$$

For the second term, $\frac{d}{dx}[3x^3y^2]$:
Apply the product rule:

$$\frac{d}{dx}[3x^3y^2] = \frac{d}{dx}(3x^3) \cdot y^2 + 3x^3 \cdot \frac{d}{dx}(y^2)$$ $$= (9x^2)y^2 + 3x^3(2y \frac{dy}{dx})$$ $$= 9x^2y^2 + 6x^3y \frac{dy}{dx}$$

For the right side, $\frac{d}{dx}[6]$:
The derivative of a constant is 0.

$$\frac{d}{dx}[6] = 0$$