Let us have function $y(x)$ and $x = \varphi(t)$. Then parametric derivative should be: $$y'_x = \dfrac{y'_t}{x'_t},$$ From where $$y'_t = y'_x \cdot x'_t = y'_x \cdot \varphi'(t).$$
The parametric derivative of the second order should be: $$y''_{xx} = (y'_x)'_x = \dfrac{(y'_x)'_t}{x'_t} = \dfrac{\left(\dfrac{y'_t}{x'_t}\right)'_t}{x'_t} = \dfrac{\dfrac{y''_{tt} \cdot x'_t - y'_t \cdot x''_{tt}}{(x'_t)^2}}{x'_t} = \dfrac{y''_{tt} \cdot x'_t - y'_t \cdot x''_{tt}}{(x'_t)^3},$$ From where $$y''_{tt} = \dfrac{y''_{xx} \cdot (x'_t)^3 + y'_t \cdot x''_{tt}}{x'_t} = \dfrac{y''_{xx} \cdot (\varphi'(t))^3 + y'_t \cdot \varphi''(t)}{\varphi'(t)}.$$
Is it correct? If not, what should I do to get $y''_{tt}$?
