Does existence and continuity of partial derivatives imply differentiability in Normed Vector Spaces? I'm wondering if the 'Normed Vector Space version' of the following theorem holds:
Theorem: let $A\subseteq \mathbb{R}^n$ be open and let $f:A\to \mathbb{R}^m$ have continuous partial derivatives $\partial f_i/\partial x_j$ on $A$. Then $f$ is differentiable on $A$.

In particular, the total derivative can be extended to NVS by considering the Fréchet Derivative. If we extend the definition of partial derivatives as follows:
Throughout the post let $V$ and $W$ be NVSs, with $V'\subseteq V$ open and $f$ a function $V'\to W$.
Definition: given $a\in V'$ and a vector $v\in V$, we define (assuming the limit exists)
$$\frac{\partial f}{\partial v}(a) := \lim_{t\rightarrow 0}\frac{f(a+tv)-f(a)}{t},$$
and, in the case $v=e_i$ for a basis vector $e_i$,
$$\frac{\partial f}{\partial x_i}(a) := \frac{\partial f}{\partial e_i}(a).$$
then, is either of the following statements true?
Theorem (?): if $f:V'\to W$ has continuous partial derivatives $\partial f/\partial v$ (for any $v\in V$) on $A$, then $f$ is differentiable on $A$.
Theorem (?): let $\dim V = n$ and $\dim W = m$. If $f:V'\to W$ have continuous partial derivatives $\partial f_i/\partial x_j$ on $A$, then $f$ is differentiable on $A$.
 A: Here is an example which (if correct, please check) shows that the first conjecture is not true.
Let $V=c_0(\mathbb{N},\mathbb{R})$ endowed with the maximum norm $\|\cdot\|$ and consider $f=(f_n):V \to V$ defined as
$$
f(x)=\left(\frac{1}{n}\sin(nx_n)\right) \quad (x=(x_n)\in V).
$$
Fix any $v=(v_n) \in V$ and let $x=(x_n) \in V$.
Note that for each $n$
$$
\frac{d}{dt}\frac{1}{n}\sin(n(x_n+tv_n)) = \cos(n(x_n+tv_n))v_n,
$$
hence by the mean value theorem
$$
\frac{1}{n}\sin(n(x_n+tv_n)) - \frac{1}{n}\sin(nx_n) = \cos(n(x_n+\tau_n(t)v_n))v_n t
$$
for some $\tau_n(t)$ between $0$ and $t$, so $|\tau_n(t)| \le |t|$.
Let $\varepsilon > 0$. Choose $n_0 \in \mathbb{N}$ such that $|v_n| \le \varepsilon/3$ $(n \ge n_0+1)$.
Now
$$
\|\frac{f(x+tv)-f(x)}{t} -  (\cos(nx_n)v_n)\| = \| (\cos(n(x_n+\tau_n(t)v_n))v_n-  \cos(nx_n)v_n)\|
$$
$$
\le \max_{n=1,\dots, n_0} | \cos(n(x_n+\tau_n(t)v_n))v_n-  \cos(nx_n)v_n)| + 2\frac{\varepsilon}{3}
$$
Since $\tau_n(t) \to 0$ $(t \to 0)$ for each $n$ there is some $\delta >0$ such that
$$
\max_{n=1,\dots, n_0} | \cos(n(x_n+\tau_n(t)v_n))v_n-  \cos(nx_n)v_n)| \le \frac{\varepsilon}{3} \quad (0<|t|< \delta).
$$
Summing up
$$
\|\frac{f(x+tv)-f(x)}{t} -  (\cos(nx_n)v_n)\| \le \varepsilon \quad (0<|t|< \delta).
$$
Thus
$$
\frac{\partial f}{\partial v}(x)= (\cos(nx_n)v_n).
$$
For each fixed $v$ (w.l.o.g $v\not=0$) the function $x \mapsto (\cos(nx_n)v_n)$ is continuous on $V$: Let  $\varepsilon > 0$ and again let $n_0 \in \mathbb{N}$ be such that $|v_n| \le \varepsilon/3$ $(n \ge n_0+1)$. For $x,y \in V$ we have
$$
\|(\cos(ny_n)v_n)-(\cos(nx_n)v_n)\| = \max_{n=1,\dots, n_0}|\cos(ny_n)v_n-\cos(nx_n)v_n| + 2\frac{\varepsilon}{3}
$$
$$
\le n_0 \|x-y\| \|v\| + 2\frac{\varepsilon}{3}.
$$
If $x$ is fixed then for each $y$ with $\|x-y\| \le \varepsilon/(3n_0 \|v\|)$
we have
$$
\|(\cos(ny_n)v_n)-(\cos(nx_n)v_n)\| \le \varepsilon.
$$
Finally we show that $f$ is not Frechet differentiable in $0=(0)$: Assume that $f$ is Frechet differentiable in $0$. Then $f'(0)h=(\cos(n0)h_n)=(h_n)$ $(h \in V)$, thus $f'(0)=id_V$.
Now consider the sequence $h^{(k)} := \frac{2\pi}{k}e_k \in V$ (with $e_k=(\delta_{nk})$).
Then $f(h^{(k)})=0$ for each $k$ and $\|h^{(k)}\| \to 0$ $(k \to \infty)$.  Hence
$$
1=\frac{\|f(h^{(k)})-f(0)-h^{(k)}\|}{\|h^{(k)}\|} \to 0 \quad (k \to \infty),
$$
a contradiction.
