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

Differentiate $\sin(x^2y^3)$ using the chain rule. Let $u = x^2y^3$.

$$\frac{d}{dx}(\sin(x^2y^3)) = \cos(x^2y^3) \cdot \frac{d}{dx}(x^2y^3)$$

To find $\frac{d}{dx}(x^2y^3)$, use the product rule, treating $y$ as a function of $x$:

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

So, the derivative of the first term is:

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

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

$$\cos(x^2y^3)(2xy^3 + 3x^2y^2 \frac{dy}{dx}) + 9x^2y^2 + 6x^3y \frac{dy}{dx} = 0$$

Distribute $\cos(x^2y^3)$:

$$2xy^3 \cos(x^2y^3) + 3x^2y^2 \cos(x^2y^3) \frac{dy}{dx} + 9x^2y^2 + 6x^3y \frac{dy}{dx} = 0$$

Group terms containing $\frac{dy}{dx}$ on one side and move other terms to the other side:

$$(3x^2y^2 \cos(x^2y^3) + 6x^3y) \frac{dy}{dx} = -2xy^3 \cos(x^2y^3) - 9x^2y^2$$

Isolate $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{-2xy^3 \cos(x^2y^3) - 9x^2y^2}{3x^2y^2 \cos(x^2y^3) + 6x^3y}$$

Factor out $-xy^2$ from the numerator:

$$-xy^2(2y \cos(x^2y^3) + 9x)$$

Factor out $3x^2y$ from the denominator:

$$3x^2y(y \cos(x^2y^3) + 2x)$$

Substitute these back into the expression for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{-xy^2(2y \cos(x^2y^3) + 9x)}{3x^2y(y \cos(x^2y^3) + 2x)}$$

Cancel $x$ and $y$ from the numerator and denominator (assuming $x \neq 0$ and $y \neq 0$):

$$\frac{dy}{dx} = \frac{-y(2y \cos(x^2y^3) + 9x)}{3x(y \cos(x^2y^3) + 2x)}$$

The final answer is $\boxed{\frac{dy}{dx} = \frac{-y(2y \cos(x^2y^3) + 9x)}{3x(y \cos(x^2y^3) + 2x)}}$.