University Calculus: Early Transcendentals (3rd Edition)

At $(0,-1)$: $$\frac{d^2y}{dx^2}=-\frac{1}{4}$$
$$xy+y^2=1$$ 1) Find $dy/dx$: Differentiate both sides of the equation with respect to $x$: $$y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0$$ $$y+(x+2y)\frac{dy}{dx}=0$$ Isolate the terms with $dy/dx$ into one side and solve for $dy/dx$: $$(x+2y)\frac{dy}{dx}=-y$$ $$\frac{dy}{dx}=-\frac{y}{x+2y}$$ 2) Find $d^2y/dx^2$: Differentiate the derivative $dy/dx$ as normal, using Quotient Rule: $$\frac{d^2y}{dx^2}=\frac{d}{dx}\Big(-\frac{y}{x+2y}\Big)=-\frac{(x+2y)\frac{dy}{dx}-y(1+2\frac{dy}{dx})}{(x+2y)^2}$$ Now we can use the result $dy/dx=-y/(x+2y)$ from part 1) to do substitution here: $$\frac{d^2y}{dx^2}=-\frac{(x+2y)\Big(-\frac{y}{x+2y}\Big)-y\Big(1-\frac{2y}{x+2y}\Big)}{(x+2y)^2}$$ $$\frac{d^2y}{dx^2}=-\Bigg[\frac{-y-y+\frac{2y^2}{(x+2y)}}{(x+2y)^2}\Bigg]=\frac{2y-\frac{2y^2}{(x+2y)}}{(x+2y)^2}$$ $$\frac{d^2y}{dx^2}=\frac{\frac{2y(x+2y)-2y^2}{(x+2y)}}{(x+2y)^2}=\frac{2xy+2y^2}{(x+2y)^3}$$ - At $(0,-1)$: $$\frac{d^2y}{dx^2}=\frac{2\times0\times(-1)+2\times(-1)^2}{\Big(0+2\times(-1)\Big)^3}=\frac{0+2}{(-2)^3}=\frac{2}{-8}=-\frac{1}{4}$$