We will present a rough sketch of the proof of the chain rule. We write $\epsilon$ for the difference, instead of $h$, in order to prevent confusion between $h$ and $h(x)$.

$$\begin{array}{rcl} f'(x)&=&\displaystyle\frac{f(x+\epsilon)-f(x)}{\epsilon} \\ &&\phantom{xx}\blue{\text{definition derivative}}\\ &=&\displaystyle\frac{g(h(x+\epsilon))-g(h(x))}{\epsilon}\\ &&\phantom{xx}\blue{f(x)=g(h(x))\text{ filled in}}\\ &=&\displaystyle\frac{g(h(x+\epsilon))-g(h(x))}{\epsilon}\cdot \frac{h(x+\epsilon)-h(x)}{h(x+\epsilon)-h(x)}\\ &&\phantom{xx}\blue{\text{multiplied with }1=\frac{h(x+\epsilon)-h(x)}{h(x+\epsilon)-h(x)}}\\ &=&\displaystyle\frac{g(h(x+\epsilon))-g(h(x))}{h(x+\epsilon)-h(x)}\cdot \frac{h(x+\epsilon)-h(x)}{\epsilon}\\ &&\phantom{xx}\blue{\text{denominators swapped}}\\ \end{array}$$

By definition, we have $h'(x)=\frac{h(x+\epsilon)-h(x)}{\epsilon}$. Showing that $\frac{g(h(x+\epsilon))-g(h(x))}{h(x+\epsilon)-h(x)}$ equals $g'h(x)$ is harder. We will use, without further explanation, the fact that if $\epsilon$ gets close to zero, the difference $h(x+\epsilon)-h(x)$ also gets close to zero. This means we can look at $h(x)+\epsilon$ instead of $h(x+\epsilon)$, because if we let $\epsilon$ get arbitrarly close to $0$, this does not really matter. We now have

$$\begin{array}{rcl}\dfrac{g(h(x+\epsilon))-g(h(x))}{h(x+\epsilon)-h(x)}&=&\dfrac{g(h(x)+\epsilon)-g(h(x))}{h(x)+\epsilon-h(x)}\\ &&\phantom{xx}\blue{h(x+\epsilon)\text{ replaced by }h(x)+\epsilon}\\ &=&\dfrac{g(h(x)+\epsilon)-g(h(x))}{\epsilon}\\ &&\phantom{xx}\blue{\text{simplified}}\\ &=&g'(h(x))\\ &&\phantom{xx}\blue{\text{definition of derivative, with }h(x)\text{ as variable}} \end{array}$$

We now see that indeed $f'(x)=g'(h(x))\cdot h'(x)$ if $f(x)=g(h(x))$.