<s>[INST] If $x = a \cos t^3 , y = b \sin t^3$ then what is $d^3y/dx^3$? If  $ x = a \cos t^3 $, $ y = b \sin t^3 $, then what is $ \frac{d^3y}{dx^3} $? 
I tried doing this problem by dividing $ \frac{d^3y}{dt^3} $ by $ \frac{d^3x}{dt^3} $ and got $ \frac{b}{a} $. 
However my book says the third derivative doesn't exist. Why is this so? 
 [/INST] Given $x = a \cos t^3,y=b\sin t^3$, 
the first derivative:
$$y'_x=\frac{y_t'}{x_t'}=\frac{(b\sin t^3)'}{(a\cos t^3)'}=\frac{b\cos t^3\cdot 3t^2}{-a\sin t^3\cdot 3t^2}=-\frac{b}{a}\cot t^3,$$
which does not exist at $t=\pi k,k\in\mathbb Z$.
The second derivative:
$$y''_{xx}=\frac{(y_x')'_t}{x_t'}=\frac{(-\frac ba\cot t^3)'}{(a\cos t^3)'}=\frac{-\frac ba\cdot \left(-\frac1{\sin ^2 t^3}\right)\cdot 3t^2}{-a\sin t^3\cdot 3t^2}=-\frac{b}{a^2}\csc^3 t^3,$$
The third derivative:
$$y'''_{xxx}=\frac{(y_{xx}'')'_t}{x_t'}=\frac{(-\frac b{a^2}\csc^3 t^3)'}{(a\cos t^3)'}=\\
=\frac{-\frac b{a^2}\cdot 3\csc ^2t^3\cdot (-\cot t^3)\cdot \csc t^3\cdot 3t^2}{-a\sin t^3\cdot 3t^2}=-\frac{3b}{a^3}\csc^4 t^3\cdot \cot t^3.$$
</s>