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"Using infinitesimals", as in physics, one can see that $\frac{\frac{df}{dt}}{\frac{dx}{dt}}=\frac{df}{dx}$.
How does one justify it?

Assuming both nominator and denominator exist, then $x$ is also continuous at t, thus $x(t+h)=x(t)+\delta x(h),\lim_{h\rightarrow 0}\delta x(h)=0$.
For $f(x(t)),x(t)$ we conclude:
$\frac{\lim_{h\rightarrow 0}\frac{f(x(t+h))-f(x(t))}{h}}{\lim_{h\rightarrow 0}\frac{x(t+h)-x(t)}{h}}=\lim_{h\rightarrow 0}\frac{\frac{f(x(t+h))-f(x(t))}{h}}{\frac{x(t+h)-x(t)}{h}}=\lim_{h\rightarrow 0}\frac{f(x(t+h))-f(x(t))}{x(t+h)-x(t)}=\lim_{\delta x\rightarrow 0}\frac{f(x+\delta x))-f(x)}{x+\delta x-x}=\lim_{\delta x\rightarrow 0}\frac{f(x+\delta x))-f(x)}{\delta x}=\frac{df}{dx}$

Have I missed anything?

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    This is simply the chain rule... just expressed differently... multiply both sides by ${dx \over dt}$ and you get the chain rule. And that has a concrete mathematical proof. – 2017-02-20
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    Be careful,as in some cases this has a negative sign,when finding the derivative of two varaibles of a 2 variable function,there is a negative sign,see:https://en.m.wikipedia.org/wiki/Chain_rule – 2017-02-20
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    You are assuming that the h's in the two limits are the same h. They are actually independent, so you need to be much more careful in your manipulations. – 2017-02-20

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