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