Although you already have a couple of answers to your question I would like to add something which you might find clarifying. As other users said, in your case, you should first evaluate the inner limit (that is, if possible, obtain a general expression for f'(x) in a neighbourhood of $a$), and then take the limit for $x \rightarrow a$.
What I wanted to point out is that, even if you assume that $f$ is differentiable at $x=a$ this is not enough to conclude that
$(1)$ \lim_{x\to a}\Big(\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\Big)= \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = f'(a).
Take for example $f(x) = \begin{cases} x^2\sin(\frac{1}{x}) \ \text{if} \ x \neq 0 \newline 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\ x=0\end{cases}$
You can check that $f$ is differentiable at $x=0$ (through the definition of derivative) but if you compute the derivative f'(x) for $x \neq 0$ and take the limit \lim_{x \rightarrow 0 }f'(x) the you will notice that such limit doesn't exist. In other words $f$ is differentiable at $x = 0$ (and in all of $\mathbb{R}$), but f' isn't continuous at $x=0$. However if the limit \lim_{x \rightarrow 0 }f'(x) did exist, you would be able to conclude that $f$ is differentiable at $x=0$, in other words it is possible to prove the following
Theorem if $f: \mathbb{R} \rightarrow \mathbb{R}$ is differentiable in a neighbourhood of $a \in \mathbb{R}$, and \lim_{x \rightarrow a }f'(x) exists and is finite, then $f$ is differentiable at $x=a$ and \lim_{x \rightarrow a }f'(x)= f'(a). (Analogous versions hold if the limit isn't finite)
Concluding $(1)$ without any particular reason (such as the hypothesis of the theorem) is very tempting, but it is also a common mistake in calculus. Hope this helps.