Differentiate the equation (x2y3) + 3x2y2 = 6 with respect to x and solve for (dy)/(dx).

The given equation is $\sin(x^2y^3) + 3x^2y^2 = 6$.
We need to apply the chain rule and product rule for differentiation. Remember that $y$ is a function of $x$, so $\frac{d}{dx}(y^n) = ny^{n-1}\frac{dy}{dx}$.

Differentiating the first term, $\sin(x^2y^3)$:
Using the chain rule, $\frac{d}{dx}(\sin(u)) = \cos(u) \frac{du}{dx}$, where $u = x^2y^3$.
To find $\frac{du}{dx} = \frac{d}{dx}(x^2y^3)$, we use the product rule: $\frac{d}{dx}(fg) = f'g + fg'$.
Let $f = x^2$ and $g = y^3$.
$f' = \frac{d}{dx}(x^2) = 2x$.
$g' = \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$.
So, $\frac{d}{dx}(x^2y^3) = (2x)(y^3) + (x^2)(3y^2 \frac{dy}{dx}) = 2xy^3 + 3x^2y^2 \frac{dy}{dx}$.
Therefore, $\frac{d}{dx}(\sin(x^2y^3)) = \cos(x^2y^3)(2xy^3 + 3x^2y^2 \frac{dy}{dx})$.

Differentiating the second term, $3x^2y^2$:
Using the product rule for $x^2y^2$:
Let $f = x^2$ and $g = y^2$.
$f' = \frac{d}{dx}(x^2) = 2x$.
$g' = \frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$.
So, $\frac{d}{dx}(3x^2y^2) = 3 \left( (2x)(y^2) + (x^2)(2y \frac{dy}{dx}) \right) = 3(2xy^2 + 2x^2y \frac{dy}{dx}) = 6xy^2 + 6x^2y \frac{dy}{dx}$.

Differentiating the constant term, 6:
$\frac{d}{dx}(6) = 0$.

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

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

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

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

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

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

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

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

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

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

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

Cancel common factors $x$ and $y$:

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

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