Let $z = 1/x$ and $y = f(z)$, find $\dfrac{d^2y}{dx^2}$
So the answer was $\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} y}{\mathrm{d} z}\frac{\mathrm{d}z }{\mathrm{d}x }$
Where $\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} y}{\mathrm{d} z}\frac{\mathrm{d}z }{\mathrm{d}x } =-\frac{\mathrm{d} y}{\mathrm{d} z}\frac{1}{x^2} = -\frac{\mathrm{d} y}{\mathrm{d} z}z^2$
$\dfrac{d^2y}{dx^2} = \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{\mathrm{d} y}{\mathrm{d} z}\frac{\mathrm{d}z }{\mathrm{d}x }\right ) = \frac{\mathrm{d} ^2y}{\mathrm{d} z^2}\left (\frac{\mathrm{d} z}{\mathrm{d} x} \right )^2 + \frac{\mathrm{d} y}{\mathrm{d} z}\frac{\mathrm{d} ^2z}{\mathrm{d} x^2} $
Could someone explain to me how on earth did $\frac{\mathrm{d} ^2y}{\mathrm{d} z^2}\left (\frac{\mathrm{d} z}{\mathrm{d} x} \right )^2$ appear?
EDIT: I am going to show what I did.
$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{\mathrm{d} y}{\mathrm{d} z}\frac{\mathrm{d}z }{\mathrm{d}x }\right ) = \frac{d}{dx}\left(\frac{dy}{dz}\right) \frac{dz}{dx}+\frac{d ^2z}{dx^2}\frac{dy}{dz} = \frac{d}{dx}\left(\frac{-1}{z^2} \frac{dy}{dx}\right) \frac{dz}{dx}+\frac{d ^2z}{dx^2}\frac{dy}{dz}$
Basically I don't understand how $\frac{d}{dx}\left(\frac{-1}{z^2} \frac{dy}{dx}\right)$ could turn into $\frac{\mathrm{d} ^2y}{\mathrm{d} z^2}\left (\frac{\mathrm{d} z}{\mathrm{d} x} \right )^2$
In fact I got $\frac{d}{dx}\left(\frac{-1}{z^2} \frac{dy}{dx}\right) = 2z^{-3}\frac{\mathrm{d} z}{\mathrm{d} x}\frac{\mathrm{d} y}{\mathrm{d} x} - \frac{1}{z^2}\frac{\mathrm{d} ^2 y}{\mathrm{d} x^2}$