Let $f(t,x)$ be a bounded and continuous function on $\mathbb{R}_t \times \mathcal{U}$ where $\mathcal{U}$ is an open neighborhood of $0 \in \mathbb{C}_x$. Moreover, assume that for each fixed $t$, $f$ is analytic in $\mathcal{U}$ with expansion of the form $$ f(t,x) = \sum_{n \geq 2} {a_n(t) x^n}. $$
Show that there is a small enough $\varepsilon > 0$ such that for some $C > 0$, $$ \left| f(t,x) \right| \leq C \varepsilon^2 \text{ whenever } |x| < \varepsilon.$$
Note: My main issue is to how deal with the coefficients $a_n(t)$.
As often is the case, writing the coefficients $a_n(t)$ explicitly as integrals helps here:
$$\begin{align} f(t,x) &= \frac{1}{2\pi i}\int_{\lvert\zeta\rvert = r} \frac{f(t,\zeta)}{\zeta-x}\,d\zeta\\ &= \frac{1}{2\pi i}\int_{\lvert\zeta\rvert = r} \frac{f(t,\zeta)}{\zeta}\frac{1}{1-\frac{x}{\zeta}}\,d\zeta\\ &= \frac{1}{2\pi i}\int_{\lvert\zeta\rvert = r} \frac{f(t,\zeta)}{\zeta} \sum_{n=0}^\infty \left(\frac{x}{\zeta}\right)^n\,d\zeta\\ &= \sum_{n=0}^\infty \underbrace{\frac{1}{2\pi i}\int_{\lvert\zeta\rvert = r} \frac{f(t,\zeta)}{\zeta^{n+1}}\,d\zeta}_{a_n(t)}\cdot x^n, \end{align}$$
where $r > 0$ is small enough that $\{ \zeta\in \mathbb{C} : \lvert\zeta\rvert \leqslant r\} \subset \mathcal{U}$, and we only consider $\lvert x\rvert < r$. The interchange of summation and integration is then legitimate by the uniform convergence of the series on the circle of integration, and we obtain the estimate
$$\lvert a_n(t)\rvert \leqslant \frac{M}{r^n}$$
for all $n$ with a bound $M$ on $\lvert f(t,\zeta)\rvert$.
By assumption, $a_0(t) \equiv a_1(t) \equiv 0$, so
$$\lvert f(t,x)\rvert \leqslant M \sum_{n=2}^\infty \left(\frac{\lvert x\rvert}{r}\right)^n = \frac{M}{r^2}\frac{1}{1-\frac{\lvert x\rvert}{r}}\cdot \lvert x\rvert^2 \leqslant \frac{2 M}{r^2}\lvert x\rvert^2$$
for all $x$ with $\lvert x\rvert \leqslant r/2$.