Prove that $\frac{d^2x}{dy^2}$ equals $-\frac{d^2y}{dx^2}\left(\frac{dy}{dx}\right)^{-3}$ The actual question is :
$$\frac{d^2x}{dy^2}$$ equals :
$$1.   \frac{d^2y}{dx^2}^{-1}$$
$$2.   -\left(\frac{d^2y}{dx^2}\right)^{-1}\left(\frac{dy}{dx}\right)^{-3}$$
$$3.\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-2}$$
$$4.-\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-3}$$
I know that the answer is 4 , in fact the solution given is as follows :
$d^2x/dy^2=-(d^2y/dx^2)(dy/dx)^{-2}(dx/dy)=$ option 4
I don't understand this.From where did the minus sign come?   Also I don't understand how the above steps are true?
This is the first time I'm solving a problem like this . Please help.
 A: Recalling that $\dfrac{dx}{dy} = \dfrac{1}{\dfrac{dy}{dx}}$, we see that 
$$\frac{d^2x}{dy^2} = \frac{d}{dy}\left(\frac{dx}{dy}\right) = \frac{d}{dy}\left(\frac{1}{\dfrac{dy}{dx}}\right)$$
Now, by quotient rule, we have that
$$\begin{aligned}\frac{d}{dy}\left(\frac{1}{\dfrac{dy}{dx}}\right) &= \frac{-\dfrac{d}{dy}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dy}{dx}\right)^2}\\ &= -\frac{\dfrac{dx}{dy}\cdot\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dy}{dx}\right)^2} \\ &= -\dfrac{\dfrac{dx}{dy}\cdot\dfrac{d^2y}{dx^2}}{\left(\dfrac{dy}{dx}\right)^2} \\ &= -\frac{\dfrac{d^2y}{dx^2}}{\left(\dfrac{dy}{dx}\right)^3}\end{aligned}$$
which is option 4.
A: Assuming that all functions are defined properly, the inverse function theorem gives
$$ \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$$
Using the quotient rule and the chain rule you get
$$ \frac{d}{dy}\left(\frac{dx}{dy}\right)
=\frac{d}{dy}\left(\frac{1}{\frac{dy}{dx}}\right)
=\frac{d}{dx}\left(\frac{1}{\frac{dy}{dx}}\right)\times \frac{dx}{dy}\\
= \frac{-\frac{d 1}{dx}\times \frac{dy}{dx}-1\times\frac{d^2 y}{dx^2}}{\left({\frac{dy}{dx}}\right)^2}\times\frac{dx}{dy}
= \frac{-\frac{d^2 y}{dx^2}}{\left({\frac{dy}{dx}}\right)^2}\times\frac{dx}{dy}
= \frac{-\frac{d^2 y}{dx^2}}{\left({\frac{dy}{dx}}\right)^2}\times\frac{1}{\frac{dy}{dx}}
= -\frac{\frac{d^2 y}{dx^2}}{\left({\frac{dy}{dx}}\right)^3}$$
