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

Step 1: Differentiate both sides of the equation with respect to $x$.
The given equation is $\sin(x^2y^3) + 3x^3y^2 = 6$.
We apply the derivative operator $\frac{d}{dx}$ to both sides:

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

Step 2: Differentiate each term.
For the first term, $\frac{d}{dx}(\sin(x^2y^3))$, we use the chain rule. 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 $g = y^3$.
$\frac{df}{dx} = 2x$.
$\frac{dg}{dx} = 3y^2 \frac{dy}{dx}$ (using the chain rule for $y^3$).
So, $\frac{du}{dx} = (2x)(y^3) + (x^2)(3y^2 \frac{dy}{dx}) = 2xy^3 + 3x^2y^2 \frac{dy}{dx}$.
Therefore, $\frac{d}{dx}(\sin(x^2y^3)) = \cos(x^2y^3) (2xy^3 + 3x^2y^2 \frac{dy}{dx})$.

For the second term, $\frac{d}{dx}(3x^3y^2)$, we use the product rule, where $f = 3x^3$ and $g = y^2$.
$\frac{df}{dx} = 9x^2$.
$\frac{dg}{dx} = 2y \frac{dy}{dx}$ (using the chain rule for $y^2$).
So, $\frac{d}{dx}(3x^3y^2) = (9x^2)(y^2) + (3x^3)(2y \frac{dy}{dx}) = 9x^2y^2 + 6x^3y \frac{dy}{dx}$.

For the right side, $\frac{d}{dx}(6) = 0$.

Step 3: Substitute the differentiated terms back into the equation.

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

Step 4: Expand and rearrange the equation to isolate $\frac{dy}{dx}$.

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

Group terms containing $\frac{dy}{dx}$ on one side and other terms on the other 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 $\frac{dy}{dx}$:

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

Step 5: Solve for $\frac{dy}{dx}$.

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

Step 6: Simplify the expression by factoring common terms from the numerator and denominator.
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 $\frac{dy}{dx}$:

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

Assuming $x \neq 0$ and $y \neq 0$, we can cancel $xy$ from the numerator and denominator:

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