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? 

We can in fact obviate the parameterisation altogether. Since $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $\frac{x}{a^2}+\frac{y}{b^2}\frac{dy}{dx}=0$ so $\frac{dy}{dx}=-\frac{b^2x}{a^2y}$ and $$\frac{d^2y}{dx^2}=-\frac{b^2}{a^2}\left(\frac{1}{y}-\frac{x}{y^2}\frac{dy}{dx}\right)=-\frac{b^4x^2}{a^4y^3}-\frac{b^2}{a^2y},\\\frac{d^3y}{dx^3}=-\frac{2b^4x}{a^4y^3}+\frac{3b^4x^2}{a^4y^4}\frac{dy}{dx}+\frac{b^2}{a^2y^2}\frac{dy}{dx}=-\frac{3b^4x}{a^4y^3}-\frac{3b^6x^3}{a^6y^5}=-\frac{3b^4x(a^2y^2+b^2x^2)}{a^6y^5}.$$We can simplify this further with $a^2y^2+b^2x^2=(ab)^2$ to $-\frac{3b^6x}{a^4y^5}$, as I mentioned in a comment on another answer.
