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? 
 A: dividing $\frac{d^3y}{dt^3}$ by $\frac{d^3x}{dt^3}$ never gives $\frac{d^3y}{dx^3}$. Actual rule is: $$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)$$
$$=\frac{x'\frac{d}{dx}(y')-y'\frac{d}{dx}(x')}{(x')^2}$$
Where $x'=\frac{dx}{dt}$ and $y'=\frac{dy}{dt}$
Now $\frac{d^3y}{dx^3}=\frac{d}{dx}\left(\frac{d^2y}{dx^2}\right)$. Use the aobve calculation here to obtain final result.
A: 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:
$$y'''_{xxx}=\frac{(y_{xx}'')'_t}{x_t'}=\frac{(-\frac b{a^2}\csc^3 t^3)'}{(a\cos t^3)'}=\\
=\frac{-\frac b{a^2}\cdot 3\csc ^2t^3\cdot (-\cot t^3)\cdot \csc t^3\cdot 3t^2}{-a\sin t^3\cdot 3t^2}=-\frac{3b}{a^3}\csc^4 t^3\cdot \cot t^3.$$
A: 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.
