If $u=g(x)$ define $\mathrm{du}/\mathrm{dx}$ as: the limit as $\delta x\to 0$ of:

$g(x+\delta x)-g(x))/\delta x$

You should have seen this definition before. The key is the precise meaning of 'limit'. Here it means that, for a given value of $x$, the above expression can be made as close as we like to $x$. (This is, roughly, the precise definition of 'limit')

More precisely, given $\epsilon >0$, we can find $\delta x$ such that the above expression is within $\epsilon$ of $\mathrm{du}/\mathrm{dx}$.

From now on I'll write $g\text{'}(x)$ for $\mathrm{du}/\mathrm{dx}$ etc.

Ok, now for the proof:

Let $u=g(x)$, $y=f(u)$. We want to prove that

$\mathrm{dy}/\mathrm{dx}=(\mathrm{dy}/\mathrm{du}).(\mathrm{du}/\mathrm{dx})$

Let $\delta x$ be a change in $x$, and $\delta u$ and $\delta y$ be the corresponding changes in $u$ and $y$.

By playing around with the limit expression above, we get:

$\delta u=(g\text{'}(x)+\epsilon )\delta x$

where $\epsilon$ depends on $x$ and $\delta x$ and $\to 0$ as $\delta x\to 0$.

Similarly for $y$:

$\delta y=(f\text{'}(u)+\epsilon \text{'})\delta u$

where, as before, $\epsilon \text{'}\to 0$ as $\delta u\to 0$.

Combining this with the expression for $\delta u$, we have:

$\delta y=(f\text{'}(u)+\epsilon \text{'})(g\text{'}(x)+\epsilon )\delta x$

NOTE: This expression is EXACT, once $\epsilon$ and $\epsilon \text{'}$ have been found, and these are uniquely defined by the value of $\delta x$.

So now divide both sides by $\delta x$ and let $\delta x\to 0$. We get:

$\mathrm{dy}/\mathrm{dx}=(\mathrm{dy}/\mathrm{du}).(\mathrm{du}/\mathrm{dx})$