$\frac{dy}{dx}$ Vs. $\dfrac{1}{\frac{dx}{dy}}$ Today, my friend gave me a question which is stated below:

On the curve $x^3 = 12y$, the abscissa changes at a faster rate than the ordinate. Then find the interval in which $x$ belongs to.

I did it as follows:

Differentiating $x^3 = 12y$ wrt $y$,
$$3x^2\frac{dx}{dy} = 12$$
$$\implies \frac{dx}{dy} = \frac{12}{3x^2}\quad, x \not = 0$$
$$\implies \frac{dx}{dy} = \frac{4}{x^2}\quad, x \not = 0$$
Now $\frac{dx}{dy}$ must be greater than $1$ so that the rate of change of abscissa is greater than rate of change of ordinate. So,
$$\dfrac{4}{x^2} > 1\quad, x \not =  0\implies \boxed{x \in (-2, 2) - \{0\}}$$
which is correct according to our book.

Whereas, my friend solved it as follows:

Differentiating $x^3 = 12y$ wrt $x$,
$$\implies 3x^2 = 12 \dfrac{dy}{dx}$$
$$\implies \dfrac{3 x^2}{12} = \dfrac{dy}{dx}$$
$$\implies \dfrac{x^2}{4} = \dfrac{dy}{dx}$$
Now, $\frac{dy}{dx}$ must be less than $1$ so that the abscissa changes at a faster rate than the ordinate. So,
$$\frac{x^2}{4} < 1 \implies\boxed{ x \in (-2, 2)}$$

What's the mistake? Whose answer is correct?
 A: 
For $$x^3 = 12y,$$ as $x$ or $y$ approaches $0,$ $\dfrac{\mathrm dx}{\mathrm dy}$ tends to $\infty$ while $\dfrac{\mathrm dy}{\mathrm dx}$ tends to $0.$
Your friend's method is correct. On the other hand, your method utilises $\dfrac{\mathrm dx}{\mathrm dy}$ but doesn't otherwise try to analyse the point $(0,0);$ so, it is incomplete.
To be clear: the given exercise doesn't intrinsically require $\dfrac{\mathrm dx}{\mathrm dy},$ or even $\dfrac{\mathrm dy}{\mathrm dx},$ to be defined at every point.
A: Let $x = t$, then the curve is the set of points
$$C = \{(x, y) \in \mathbb{R}^2 | (x, y) = (t, t^3/12), t \in \mathbb{R}\}.$$
Then taking the derivative with respect to $t$ you get the vector $(1, t^2/4)$. This is the vector of the "rates of change" for abscissa and ordinate respectively. To find the region of the parameter where the first is larger than the second, we find
$$ 1 > t^2/4 \implies t \in (-2, 2).$$
Since $t=x$, this implies $x \in (-2, 2)$. Your method fails to find $x=0$ and I believe the book is incorrect. If you plot the curve it's clear that the curve is defined at $x=0$ and that the derivative at that point is zero, which is clearly less than one.
A: The answers are equivalent: in your friend's answer he has to notice that if $x = 0$ then $y=0$. Hnece $dy/dx = 0$ which is not less than $1$ and so $x = 0$ is not a solution and has to be discarted.
The difference in the two approaches is that this requirement is apparent in your solution while it has to be noticed in your friend's answer.
Hope this helps.
A: I could be wrong, but I think the last line of your friend's answer is missing something:
$$0 < \frac{x^2}{4} < 1 \implies x \in (-2,2) - \{0\}$$
This is because if you want the abscissa to change at a faster rate than the ordinate, then you need $0 < dy/dx < 1$. But then again, when $dy/dx = 0$, then the relation $x^3 = 12y$ is stationary which means the abscissa is changing faster than the ordinate... Hence $x \in (-2,2)$.
It's quite difficult to judge, so please do let me know what you think.
