#### Answer

Differentiate implicitly the given equation in both a) and b) to prove the given results.

#### Work Step by Step

a) $$x^2-y^2=1$$
To show that $dy/dx=x/y$, we will use implicit differentiation to find $dy/dx$ directly.
Differentiate both sides of the equation with respect to $x$: $$2x-2y\frac{dy}{dx}=0$$
$$2y\frac{dy}{dx}=2x$$
$$\frac{dy}{dx}=\frac{2x}{2y}=\frac{x}{y}$$
As we can see, $dy/dx$ equals $x/y$ here.
b) Next, to show that $d^2y/dx=-1/y^3$, we use implicit differentiation again on $dy/dx$ to find $d^2y/dx^2$ directly:
$$\frac{d^2y}{dx^2}=\frac{\frac{dx}{dx}\times y-\frac{dy}{dx}\times x}{y^2}=\frac{y-x\frac{dy}{dx}}{y^2}$$
Here, we replace $dy/dx=x/y$: $$\frac{d^2y}{dx^2}=\frac{y-x\times\frac{x}{y}}{y^2}=\frac{y-\frac{x^2}{y}}{y^2}$$
$$\frac{d^2y}{dx^2}=\frac{\frac{y^2-x^2}{y}}{y^2}=\frac{y^2-x^2}{y^3}$$
From the exercise, we have the equation $x^2y-y^2=1$, which means $y^2-x^2=-1$
$$\frac{d^2y}{dx^2}=\frac{-1}{y^3}$$
As we can see again, $d^2y/dx^2$ equals $-1/y^3$ here.