Step 1: Differentiate each term of the equation $\sin(x^2y^3) + 3x^3y^2 = 6$ with respect to $x$. Differentiate $\sin(x^2y^3)$ using the chain rule. Let $u = x^2y^3$. $$ \frac{d}{dx}(\sin(x^2y^3)) = \cos(x^2y^3) \cdot \frac{d}{dx}(x^2y^3) $$ To find $\frac{d}{dx}(x^2y^3)$, use the product rule, treating $y$ as a function of $x$: $$ \frac{d}{dx}(x^2y^3) = \frac{d}{dx}(x^2) \cdot y^3 + x^2 \cdot \frac{d}{dx}(y^3) $$ $$ \frac{d}{dx}(x^2y^3) = 2xy^3 + x^2(3y^2 \frac{dy}{dx}) = 2xy^3 + 3x^2y^2 \frac{dy}{dx} $$ So, the derivative of the first term is: $$ \cos(x^2y^3)(2xy^3 + 3x^2y^2 \frac{dy}{dx}) $$ Step 2: Differentiate $3x^3y^2$ using the product rule, treating $y$ as a function of $x$: $$ \frac{d}{dx}(3x^3y^2) = 3 \left( \frac{d}{dx}(x^3) \cdot y^2 + x^3 \cdot \frac{d}{dx}(y^2) \right) $$ $$ \frac{d}{dx}(3x^3y^2) = 3(3x^2y^2 + x^3(2y \frac{dy}{dx})) $$ $$ \frac{d}{dx}(3x^3y^2) = 9x^2y^2 + 6x^3y \frac{dy}{dx} $$ Step 3: Differentiate the constant term $6$: $$ \frac{d}{dx}(6) = 0 $$ Step 4: Combine the derivatives and solve for $\frac{dy}{dx}$. $$ \cos(x^2y^3)(2xy^3 + 3x^2y^2 \frac{dy}{dx}) + 9x^2y^2 + 6x^3y \frac{dy}{dx} = 0 $$ Distribute $\cos(x^2y^3)$: $$ 2xy^3 \cos(x^2y^3) + 3x^2y^2 \cos(x^2y^3) \frac{dy}{dx} + 9x^2y^2 + 6x^3y \frac{dy}{dx} = 0 $$ Group terms containing $\frac{dy}{dx}$ on one side and move other terms to the other side: $$ (3x^2y^2 \cos(x^2y^3) + 6x^3y) \frac{dy}{dx} = -2xy^3 \cos(x^2y^3) - 9x^2y^2 $$ Isolate $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{-2xy^3 \cos(x^2y^3) - 9x^2y^2}{3x^2y^2 \cos(x^2y^3) + 6x^3y} $$ Step 5: Simplify the expression by factoring out common terms from the numerator and denominator. Factor out $-xy^2$ from the numerator: $$ -xy^2(2y \cos(x^2y^3) + 9x) $$ Factor out $3x^2y$ from the denominator: $$ 3x^2y(y \cos(x^2y^3) + 2x) $$ Substitute these back into the expression for $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{-xy^2(2y \cos(x^2y^3) + 9x)}{3x^2y(y \cos(x^2y^3) + 2x)} $$ Cancel $x$ and $y$ 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 $\boxed{\frac{dy}{dx} = \frac{-y(2y \cos(x^2y^3) + 9x)}{3x(y \cos(x^2y^3) + 2x)}}$. Send me the next one 📸