We have the following relevant fact:
If $\mathscr{F}\subset \mathscr{O}(U)$ is a normal family, then $$\mathscr{F}' = \{ f' : f \in \mathscr{F}\}$$ is also a normal family.
The proof is simple, for any sequence $(g_n)$ in $\mathscr{F}'$ select a sequence $(f_n)$ in $\mathscr{F}$ such that $f_n' = g_n$ for all $n$. By the normality of $\mathscr{F}$, we can extract a compactly convergent subsequence. Without loss of generality - to save subscripting - assume the whole sequence $(f_n)$ is compactly convergent with limit $f$. For every $z_0 \in U$ choose $r > 0$ such that $D_{3r}(z_0)\subset U$. For $z \in D_r(z_0)$ we then have
$$g_n(z) = f_n'(z) = \frac{1}{2\pi i} \int_{\lvert\zeta-z_0\rvert = 2r} \frac{f_n(\zeta)}{(\zeta-z)^2}\,d\zeta.$$
Since the circle $C = \{\zeta : \lvert\zeta-z_0\rvert = 2r\}$ is compact, $f_n$ converges uniformly to $f$ on $C$, and since $\lvert \zeta-z\rvert > r$ for $(\zeta,z) \in C \times D_r(z_0)$, we have
$$\frac{f_n(\zeta)}{(\zeta-z)^2} \to \frac{f(\zeta)}{(\zeta-z)^2}$$
uniformly on $C\times D_r(z_0)$, whence
$$g_n(z) \to g(z) = \frac{1}{2\pi i} \int_{\lvert\zeta-z_0\rvert = 2r} \frac{f(\zeta)}{(\zeta-z)^2}\,d\zeta = f'(z)$$
uniformly on $D_r(z_0)$. Since $z_0$ was arbitrary, it follows that $(g_n)$ converges locally uniformly, hence compactly, on $U$.
This can be iterated, so for a normal family $\mathscr{F}$, the family $\mathscr{F}^{(k)} = \{ f^{(k)} : f \in \mathscr{F}\}$ of $k^{\text{th}}$ derivatives is again normal.
For our setting, where $\mathscr{F} = \{ f^n : n \in \mathbb{N}\}$, we need only look at the derivatives at $0$. If $\mathscr{F}$ is normal, then the sequence $\bigl((f^n)^{(k)}(0)\bigr)_{n\in \mathbb{N}}$ must be bounded for every $k$. Investigating that will lead you to a necessary and sufficient condition.
