Chain rule

Since $f$ is differentiable at $g(x)$, write $f(g(x)+k) - f(g(x)) = [f'(g(x)) + \varepsilon(k)]\,k$ for $k$ near $0$, where $\varepsilon(k)\to 0$ as $k\to 0$ (setting $\varepsilon(0)=0$). Substitute $k = g(x+h)-g(x)$, which tends to $0$ as $h\to 0$ by continuity of $g$ at $x$ (itself a consequence of differentiability), and divide by $h$: $\frac{f(g(x+h))-f(g(x))}{h} = [f'(g(x))+\varepsilon(k)]\cdot\frac{g(x+h)-g(x)}{h} \to f'(g(x))\,g'(x)$ as $h\to 0$.