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

For the first term, $\frac{d}{dx}(\sin(x^2y^3))$, we use the chain rule. Let $u = x^2y^3$.

$$\frac{d}{dx}(\sin(u)) = \cos(u) \cdot \frac{du}{dx}$$

Now, we need to find $\frac{du}{dx} = \frac{d}{dx}(x^2y^3)$ using the product rule.

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

So, the derivative of the first term is:

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

For the second term, $\frac{d}{dx}(3x^3y^2)$, we use the product rule.

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

The derivative of the right side is $\frac{d}{dx}(6) = 0$.

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

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

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

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

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

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

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

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

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

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

So,

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

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