Differentiate the equation sin(x2y3) + 3x3y2 = 6 implicitly with respect to x.

The given equation is $\sin(x^2y^3) + 3x^3y^2 = 6$.

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

For the first term, $\frac{d}{dx}[\sin(x^2y^3)]$:
Using the chain rule, $\frac{d}{dx}[\sin(u)] = \cos(u) \frac{du}{dx}$, where $u = x^2y^3$.
Now, find $\frac{du}{dx} = \frac{d}{dx}[x^2y^3]$ using the product rule.
Let $f = x^2$ and $g = y^3$. Then $\frac{df}{dx} = 2x$ and $\frac{dg}{dx} = 3y^2 \frac{dy}{dx}$.
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, the derivative of the first term is:

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

For the second term, $\frac{d}{dx}[3x^3y^2]$:
Using the product rule, let $f = 3x^3$ and $g = y^2$. Then $\frac{df}{dx} = 9x^2$ and $\frac{dg}{dx} = 2y \frac{dy}{dx}$.
So, the derivative of the second term is:

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

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

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

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

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

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

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

Factor out $3xy$ from the denominator:

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

So,

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

Cancel $xy$ from the numerator and denominator (assuming $x \neq 0$ and $y \neq 0$):

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

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