actually it is looking to be true. I wasn't quite evaluating it correctly. I'll need to toy with this a bit or maybe one of the math powerhouses will chime in with something deviously clever.
This wiki page has the derivation you need.
look at the section on higher derivatives.
Given any $\displaystyle y = f(x)$, then $\displaystyle y' = f'(x)$. Using implicit differentiation and taking the derivative of both sides with respect to $\displaystyle y$, you get:
$\displaystyle \begin{align*}\dfrac{d}{dy}(y) & = \dfrac{d}{dy}\left(f(x)\right) \\ 1 & = f'(x)\dfrac{dx}{dy} \\ \dfrac{dx}{dy} & = \dfrac{1}{y'}\end{align*}$
For the second derivative:
$\displaystyle \begin{align*}\dfrac{d}{dy}\left( \dfrac{dx}{dy} \right) & = \dfrac{d}{dy}\left( \dfrac{1}{y'} \right) \\ \dfrac{d^2x}{dy^2} & = -\dfrac{y''\dfrac{dx}{dy}}{(y')^2} \\ & = -\dfrac{y''}{(y')^3}\end{align*}$
One more time:
$\displaystyle \begin{align*}\dfrac{d}{dy}\left(\dfrac{d^2x}{dy^2 }\right) & = \dfrac{d}{dy}\left(-\dfrac{y''}{(y')^3}\right) \\ \dfrac{d^3x}{dy^3} & = -\dfrac{(y')^3y''' \dfrac{dx}{dy} - 3(y'')^2(y')^2\dfrac{dx}{dy}}{(y')^6} \\ & = \dfrac{3(y'')^2(y') - (y')^2y'''}{(y')^6} \\ & = \dfrac{3(y'')^2}{(y')^5} - \dfrac{y'''}{(y')^4}\end{align*}$