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

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

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

Here, $f = 3x^3$ and $g = y^2$.
$\frac{d}{dx}(3x^3) = 9x^2$
$\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$ (by chain rule)
So, the derivative of the second term is:

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

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

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

Factor out $-xy^2$ from the numerator:
Numerator $= -xy^2 (2y \cos(x^2y^3) + 9x)$
Factor out $3xy$ from the denominator:
Denominator $= 3xy (xy \cos(x^2y^3) + 2x^2)$
Substitute these back into the expression for $\frac{dy}{dx}$:

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

Cancel out $xy$ (assuming $x \neq 0$ and $y \neq 0$):

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

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