Find the second derivative ${{{d^2}y} \over {d{x^2}}}$ in terms of t when $x = 3 - 2{t^2}$ and $y = {1 \over t}$ This is my attempt:
$\eqalign{
  & x = 3 - 2{t^2}  \cr 
  & y = {1 \over t}  \cr 
  & {{dx} \over {dt}} =  - 4t  \cr 
  & {{dy} \over {dt}} =  - {t^{ - 2}} = {{ - 1} \over {{t^2}}}  \cr 
  & {{dy} \over {dx}} = {{ - 1} \over {{t^2}}} \times {1 \over { - 4t}} = {1 \over {4{t^3}}}  \cr 
  & {{{d^2}y} \over {d{x^2}}} = {d \over {dx}}\left( {{{dy} \over {dx}}} \right) = {{{d \over {dx}}\left( {{{dy} \over {dt}}} \right)} \over {{{dx} \over {dt}}}} = {{{d \over {dt}}\left( {{{dy} \over {dx}}} \right)} \over {{{dx} \over {dt}}}}  \cr 
  & {{{d^2}y} \over {d{x^2}}}:  \cr 
  & {d \over {dt}}\left( {{{dy} \over {dx}}} \right) =  - 1{(4{t^3})^{ - 2}} \times 12{t^2} = {{ - 12{t^2}} \over {16{t^6}}} = {{ - 3t} \over {4{t^4}}}  \cr 
  & {{dx} \over {dt}} =  - 4t  \cr 
  & {{{d^2}y} \over {d{x^2}}} = {{ - 3t} \over {4{t^4}}} \times {1 \over { - 4t}} = {{3t} \over {16{t^5}}} \cr} $

After some wrangling I've finally got that answer, which is right according to the textbook. The problem I was experiencing was knowing when to treat an expression as a function or a variable? I recognise that you have to differentiate the two differently, but I dont completely understand it. Why couldn't I differentiate the numerator of the second differential ${d \over {dt}}\left( {{{dy} \over {dx}}} \right)$ like this: 
$$\eqalign{
  & {{dy} \over {dx}} = 4{t^{ - 3}}  \cr 
  & {d \over {dt}}\left( {{{dy} \over {dx}}} \right) =  - 12{t^{ - 4}} \cr} $$
T is a variable of the function x(t), so should I not be able to differentiate it as normal variable?
I hope you have kept with me and this doesnt sound convoluted, thank you.

Recall that:
$$\frac {dy}{dx}  = \frac{dy}{dt}\cdot \frac{dt}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
So what you propose at the end, is just finding $$\dfrac{dy}{dt} = -12t^{-4}$$
But to find $d^2y/dx^2$, we need to account again for finding each the derivative of y with respect to $t$, and the derivative of $x$ with respect to $t$:
$${{{d^2}y} \over {d{x^2}}} = {d \over {dx}}\left( {{{dy} \over {dx}}} \right) = {{{d \over {dx}}\left( {{{dy} \over {dt}}} \right)} \over {{{dx} \over {dt}}}} = {{{d \over {dt}}\left( {{{dy} \over {dx}}} \right)} \over {{{dx} \over {dt}}}}$$
$${{{d^2}y} \over {d{x^2}}}: {d \over {dt}}\left( {{{dy} \over {dx}}} \right) =  - 1{(4{t^3})^{ - 2}} \times 12{t^2}$$ $$ = {{ - 12{t^2}} \over {16{t^6}}} = {{ - 3t} \over {4{t^4}}}$$
