y`_(t)=(1-t^3)`=-3t^2
[m]\frac{dy}{dx}=\frac{y`_{t}}{x`_{t}}=\frac{(-3t^2)}{-(2t)}=\frac{3}{2}t[/m]
(y`_(x))_(t)=[m](\frac{3}{2}t)`=\frac{3}{2}[/m]
[m]\frac{d^2y}{dx^2}=\frac{(y`_{x})`_{t}}{x`_{t}}=\frac{\frac{3}{2}}{-(2t)}=-\frac{3}{4t}[/m]