Answer
$$\frac{dy}{dx}=-\Big(\frac{y}{x}\Big)^{1/3}$$
$$\frac{d^2y}{dx^2}=\frac{y^{-1/3}+x^{-2/3}y^{1/3}}{3x^{2/3}}$$
Work Step by Step
$$x^{2/3}+y^{2/3}=1$$
1) Find $dy/dx$:
Differentiate both sides of the equation with respect to $x$: $$\frac{d}{dx}(x^{2/3})+\frac{d}{dx}(y^{2/3})=\frac{d}{dx}1$$ $$\frac{2}{3}x^{-1/3}+\frac{2}{3}y^{-1/3}\frac{dy}{dx}=0$$
Isolate the terms with $dy/dx$ and solve for $dy/dx$: $$\frac{2}{3}y^{-1/3}\frac{dy}{dx}=-\frac{2}{3}x^{-1/3}$$ $$\frac{dy}{dx}=-\frac{x^{-1/3}}{y^{-1/3}}=-\frac{y^{1/3}}{x^{1/3}}=-\Big(\frac{y}{x}\Big)^{1/3}$$
2) Find $d^2y/dx^2$:
Differentiate the derivative $dy/dx$ as normal, using Quotient Rule: $$\frac{d^2y}{dx^2}=\frac{d}{dx}\Big(-\frac{y^{1/3}}{x^{1/3}}\Big)=-\frac{(\frac{1}{3}y^{-2/3}\frac{dy}{dx})x^{1/3}-(\frac{1}{3}x^{-2/3}y^{1/3})}{x^{2/3}}$$
Now we can use the result $dy/dx=-y^{1/3}/x^{1/3}$ from part 1) to do substitution here:
$$\frac{d^2y}{dx^2}=\frac{\Big(\frac{1}{3}y^{-2/3}(-\frac{y^{1/3}}{x^{1/3}})\Big)x^{1/3}-(\frac{1}{3}x^{-2/3}y^{1/3})}{x^{2/3}}$$ $$\frac{d^2y}{dx^2}=-\frac{-\frac{1}{3}\frac{y^{-2/3+1/3}}{x^{1/3-1/3}}-(\frac{1}{3}x^{-2/3}y^{1/3})}{x^{2/3}}$$ $$\frac{d^2y}{dx^2}=\frac{\frac{1}{3}y^{-1/3}+\frac{1}{3}x^{-2/3}y^{1/3}}{x^{2/3}}$$ $$\frac{d^2y}{dx^2}=\frac{y^{-1/3}+x^{-2/3}y^{1/3}}{3x^{2/3}}$$
