Differentiate the equation (x2 y3) + 3x3 y2 = 6 with respect to x and solve for (dy)/(dx).

Given the equation:

$$\sin(x^2 y^3) + 3x^3 y^2 = 6$$

Differentiate each term with respect to $x$:

$$\frac{d}{dx}(\sin(x^2 y^3)) + \frac{d}{dx}(3x^3 y^2) = \frac{d}{dx}(6)$$

Let $u = x^2 y^3$. Then $\frac{d}{dx}(\sin u) = \cos u \cdot \frac{du}{dx}$.
First, find $\frac{du}{dx} = \frac{d}{dx}(x^2 y^3)$ using the product rule, where $y$ is a function of $x$.

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

Now substitute this back into the derivative of the first term:

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

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

Substitute the results from Step 2, Step 3, and Step 4 back into the equation from Step 1:

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

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

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

Factor out $\frac{dy}{dx}$:

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

Factor out common terms from the numerator and the denominator.
Numerator: $-xy^2 (2y \cos(x^2 y^3) + 9x)$
Denominator: $3x^2 y (y \cos(x^2 y^3) + 2x)$

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

Cancel out common factors $\frac{xy^2}{x^2 y} = \frac{y}{x}$:

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

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