Find dy/dx for (x2y3) + 3x3y2 = 6.

We use 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 $g = y^3$.
Then $f' = \frac{d}{dx}(x^2) = 2x$.
And $g' = \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$ (by chain rule, since $y$ is a function of $x$).
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, the derivative of the first term is:

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

We use the product rule. Let $f = 3x^3$ and $g = y^2$.
Then $f' = \frac{d}{dx}(3x^3) = 9x^2$.
And $g' = \frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$.
So, the derivative of the second term is:

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

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

Expand the first term:

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

Move terms without $\frac{dy}{dx}$ to the right side of the equation:

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

From the numerator, factor out $-xy^2$:

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

From the denominator, factor out $3x^2y$:

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

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 out the common factor $xy$:

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

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