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

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

$$\frac{d}{dx}[\sin(x^2y^3) + 3x^3y^2] = \frac{d}{dx}[6]$$ $$\frac{d}{dx}[\sin(x^2y^3)] + \frac{d}{dx}[3x^3y^2] = 0$$

Let $u = x^2y^3$. Then $\frac{d}{dx}[\sin(u)] = \cos(u) \frac{du}{dx}$.
To find $\frac{du}{dx} = \frac{d}{dx}[x^2y^3]$, we use the product rule, where $f = x^2$ and $g = y^3$.
$\frac{d}{dx}[x^2y^3] = \frac{d}{dx}[x^2] \cdot y^3 + x^2 \cdot \frac{d}{dx}[y^3]$

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

So, the derivative of the first term is:

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

Here, $f = 3x^3$ and $g = y^2$.
$\frac{d}{dx}[3x^3y^2] = \frac{d}{dx}[3x^3] \cdot y^2 + 3x^3 \cdot \frac{d}{dx}[y^2]$

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

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

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 $\frac{dy}{dx}$ from the left side:

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

Factor $-xy^2$ from the numerator:
Numerator $= -xy^2(2y \cos(x^2y^3) + 9x)$
Factor $3xy$ from the denominator:
Denominator $= 3xy(xy \cos(x^2y^3) + 2x^2)$
Now, substitute these factored forms 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 $xy$ from the numerator and denominator (assuming $