TITLE: When can't $dy/dx$ be used as a ratio/fraction?
QUESTION [5 upvotes]: By searching this question, I found this:
Can I ever go wrong if I keep thinking of derivatives as ratios?
However, the answers don't have what I'm looking for! (Edit: Meaning, a counterexample. There is one involving partial derivatives, but then the only difference has to do with signs, which means that $dy/dx$ can still be interpreted as a ratio. Thanks, fuglege)
So long as you treat $dx$ as $dx$ (meaning one "object", that is, not d times x), I still have yet to see an example where using the differentials as a fraction yields an incorrect answer (and I have watched almost all the khan academy videos on calculus and read quite a bit out of the ubiquitous Stewart calculus book!)
Note that I am not asking about non-standard analysis. I have found an online textbook and I am currently reading it. I am only trying to find a counterexample to using the differentials as a fraction.
Thanks very much

REPLY [2 votes]: The problem comes when you hit the second-order differentials.  When using second-order differentials in the standard notation, they DO NOT cancel.  
Example: Let's say that $y = x^3$ and $x = t^2$.  Find the second derivative of the first one:
$$y = x^3 \\
\frac{dy}{dx} = 3x^2 \\
\frac{d^2y}{dx^2} = 6x$$
Now, take the derivative of $x = t^2$.  You get:
$$\frac{dx}{dt} = 2t$$.
Now, let's try to cancel by squaring the second one:
$$\frac{d^2y}{dx^2}\cdot\left(\frac{dx}{dt}\right)^2 = 6x\cdot (2t)^2 \\
\frac{d^2y}{dx^2}\cdot\frac{dx^2}{dt^2} = 6(t^2) (2t^2) \\
\frac{d^2y}{dt^2} = 24t^4$$
The problem?  Let's do it again, but substitute at the beginning.  Since $y = x^3$ and $x = t^2$, that means that $y = (t^2)^3 = t^6$.  What's the second derivative?
$$\frac{dy}{dt} = 6t^5 \\
\frac{d^2y}{dt^2} = 30t^4$$
Uh oh!  Contradiction!
Now, the problem isn't with the concept of treating differentials as fractions, but our implementation of it. See Extending the Algebraic Manipulability of Differentials.  When you take the derivative of the derivative, if you are treating $\frac{dy}{dx}$ as a fraction, then the derivative of that fraction should use the quotient rule.  Therefore, the notation of the second derivative should be:
$$\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$
Using this notation, then the higher-order differentials become algebraically manipulable again.  I could demonstrate it for you, but it is pretty messy.  See the paper if you want to see it happen.