1. The problem is to find the derivative $y'$ for the equation $$\frac{x}{y^3} = 1$$ using two methods: (a) solving explicitly for $y$ and then differentiating, and (b) using implicit differentiation. 2. First, solve the equation for $y$ explicitly: $$\frac{x}{y^3} = 1 \implies y^3 = x \implies y = x^{\frac{1}{3}}$$ 3. Differentiate $y = x^{\frac{1}{3}}$ with respect to $x$ using the power rule: $$y' = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}}$$ 4. Now, use implicit differentiation on the original equation: Start with: $$\frac{x}{y^3} = 1$$ Rewrite as: $$x = y^3$$ Differentiate both sides with respect to $x$: $$1 = 3y^2 \frac{dy}{dx} = 3y^2 y'$$ 5. Solve for $y'$: $$y' = \frac{1}{3y^2}$$ 6. Check that the two derivatives are equivalent by substituting $y = x^{\frac{1}{3}}$ into the implicit derivative: $$y' = \frac{1}{3 (x^{\frac{1}{3}})^2} = \frac{1}{3 x^{\frac{2}{3}}}$$ This matches the explicit derivative found in step 3. Final answer: $$y' = \frac{1}{3x^{\frac{2}{3}}}$$