Let me assume that
$
f(t,x)=\sum_{k\ge 0} f_k(x) t^k, \quad \vert x\vert \le 1, \quad \vert t\vert < 1,
$
with $f_k$ Lipschitz-continuous with an $L^\infty$ norm on $\vert x\vert \le 1$ bounded above by 1 and $\Vert f'_k\Vert_{L^\infty}\le C_0 R^k$.
Then each $f_k$ is a.e. differentiable, i.e. $\forall k, \exists D_k, \vert D_k^c\vert=0$, $f_k$ is differentiable on $D_k$ (the complement is taken in the unit ball in $x$). Defining
$
D=\cap_{k\ge 0} D_k,
$
gives that $\vert D^c\vert=0$ and all $f_k$ are differentiable on $D$. We have
in the distribution sense on the open cube $\{\vert x\vert < 1, \vert t\vert <R^{-1}\}$
$$
\frac{\partial f}{\partial x} (t,x)=\sum_{k\ge 0}f'_k(x) t^k,
$$
and for $h\not=0$,
\begin{multline}
\frac1h\bigl(f(t,x+h)-f(t,x)\bigr)=\sum_{k\ge 0}\frac1h\bigl(f_k(x+h)-f_k(x)\bigr) t^k
\\=\sum_{k\ge 0}\frac1h\int_{x}^{x+h} \bigl(f'_k(y)-f'_k(x)\bigr) dy t^k
+\sum_{k\ge 0} f'_k(x)t^k.
\end{multline}
Thanks to the Lebesgue Differentiation Theorem, each $\frac1h\int_{x}^{x+h} \bigl(f'_k(y)-f'_k(x)\bigr) dy$ goes to zero with $h$ for $x\in D$. We have also (for $h>0$, $\vert t\vert<1/R$)
$$
\sum_{k\ge N}\frac1{h}\int_{x}^{x+h} \bigl\vert f'_k(y)-f'_k(x)\bigr\vert dy \vert t\vert^k\le
\sum_{k\ge N} 2\Vert f'_k\Vert_{L^\infty}\vert t\vert^k\le 2C_0(R\vert t\vert)^N\frac{1}{1-R\vert t\vert}.
$$
We get that for all $N\ge 0$,
$$
\limsup_{h\rightarrow 0}\left\vert \frac1h\bigl(f(t,x+h)-f(t,x)\bigr) -\sum_{k\ge 0}f'_k(x) t^k\right\vert
\le \frac{ 2C_0(R\vert t\vert)^N}{1-R\vert t\vert},
$$
which implies for $\vert t\vert<1/R$, $x\in D$,
$$
\lim_{h\rightarrow 0} \frac1h\bigl(f(t,x+h)-f(t,x)\bigr) =\sum_{k\ge 0}f'_k(x) t^k.
$$