suppose that x and y are related by the given equation and use implicit differentiation to determine dy/dx…

Explanation:

Step1: Differentiate both sides

Differentiate $x^{3}y^{3}$ using product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x^{3}$ and $v = y^{3}$. The derivative of the right - hand side of the equation $\frac{d}{dx}(6)=0$. $\frac{d}{dx}(x^{3}y^{3})=\frac{d}{dx}(6)$ $(3x^{2})y^{3}+x^{3}(3y^{2}\frac{dy}{dx}) = 0$

Step2: Isolate $\frac{dy}{dx}$

First, factor out common terms: $3x^{2}y^{3}+3x^{3}y^{2}\frac{dy}{dx}=0$ $3x^{2}y^{2}(y + x\frac{dy}{dx})=0$. Since we want to solve for $\frac{dy}{dx}$, we can rewrite the equation as: $3x^{3}y^{2}\frac{dy}{dx}=- 3x^{2}y^{3}$ Then divide both sides by $3x^{3}y^{2}$ (assuming $x\neq0$ and $y\neq0$): $\frac{dy}{dx}=-\frac{y}{x}$

Answer:

$-\frac{y}{x}$