Step 1: Differentiate both sides of the equation $\sin(x^2y^3) + 3x^3y^2 = 6$ with respect to $x$. $$ \frac{d}{dx}[\sin(x^2y^3) + 3x^3y^2] = \frac{d}{dx}[6] $$ Step 2: Differentiate each term. For the first term, $\frac{d}{dx}[\sin(x^2y^3)]$: Apply the chain rule. Let $u = x^2y^3$. Then $\frac{d}{dx}[\sin(u)] = \cos(u) \frac{du}{dx}$. To find $\frac{du}{dx} = \frac{d}{dx}[x^2y^3]$, apply 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: $$ \cos(x^2y^3) (2xy^3 + 3x^2y^2 \frac{dy}{dx}) = 2xy^3 \cos(x^2y^3) + 3x^2y^2 \cos(x^2y^3) \frac{dy}{dx} $$ For the second term, $\frac{d}{dx}[3x^3y^2]$: Apply the product rule: $$ \frac{d}{dx}[3x^3y^2] = \frac{d}{dx}(3x^3) \cdot y^2 + 3x^3 \cdot \frac{d}{dx}(y^2) $$ $$ = (9x^2)y^2 + 3x^3(2y \frac{dy}{dx}) $$ $$ = 9x^2y^2 + 6x^3y \frac{dy}{dx} $$ For the right side, $\frac{d}{dx}[6]$: The derivative of a constant is $0$. $$ \frac{d}{dx}[6] = 0 $$ Step 3: Combine the differentiated terms into the equation. $$ 2xy^3 \cos(x^2y^3) + 3x^2y^2 \cos(x^2y^3) \frac{dy}{dx} + 9x^2y^2 + 6x^3y \frac{dy}{dx} = 0 $$ Step 4: Group terms containing $\frac{dy}{dx}$ on one side and other terms on the other side. $$ 3x^2y^2 \cos(x^2y^3) \frac{dy}{dx} + 6x^3y \frac{dy}{dx} = -2xy^3 \cos(x^2y^3) - 9x^2y^2 $$ Step 5: Factor out $\frac{dy}{dx