I know that if $E$ and $F$ are finite-dimensional Banach spaces and $f:E\to F$ is differentiable, then $\mathrm{d}f$ does not depend on the choice of norms in $E$ and $F$ (since all the norms are equivalent). However, that does not seem to be the case when $E$ or $F$ are infinite-dimensional.
I would like to find an explicit exemple of infinite-dimensional Banach spaces $E,F$ and a function $f:E\to F$ such that $\mathrm{d}f$ is different when we choose one norm for $E,F$ when compared to another norm in $E,F$. It would be nice too to see an exemple where the $f$ is differentiable for one norm but not differentiable for another.
First, note that if you are only interested in what the differential $D f$ is (and not in whether $f$ is differentiable to begin with), then the norm on $E$ is immaterial. Indeed, you have $$ D f (x_0) v = \lim_{t \to 0} \frac{f(x_0 + t v) - f(x_0)}{t} \, , $$ which is a limit in $F$, completely independent of the topology on $E$.
Next, let us see that if $\|\cdot\|$ and $|\cdot|$ are two norms on $F$ such that $F$ is complete for any of the two norms and such that the two norms are not equivalent, then there is a function $g : (-1,1) \to F$ such that $g$ is differentiable at $0$ for both choices of the norm, but with different differentials.
For brevity, let us write $F_1$ for the vector space $F$ equipped with the norm $\|\cdot\|$, and $F_2$ for the vector space $F$ equipped with the norm $|\cdot|$.
The following lemma will be crucial:
Lemma. There is a sequence $(x_n)_n$ in $F$ and $x,y \in F$ with $x \neq y$ such that $\|x_n - x\| \to 0$ and $|x_n - y| \to 0$.
Proof. Assume towards a contradiction that the claim fails. Consider the linear operator $$ T : F_1 \to F_2, x \mapsto x \, . $$ I claim that $T$ has closed graph. Indeed, let $(x_n, y_n)_{n \in \mathbb{N}} \subset \mathrm{graph}(T) \subset F_1 \times F_2$ be a sequence in the graph of $T$, with $(x_n, y_n) \to (x,y) \in F_1 \times F_2$. Since $(x_n, y_n) \in \mathrm{graph}(T)$, we have $x_n = y_n$ for all $n \in \mathbb{N}$, and thus $\|x_n - x\| \to 0$ and $|x_n - y| = |y_n - y| \to 0$. Since we assume that the claim is false, we must have $x = y$, and thus $(x,y) \in \mathrm{graph}(T)$, as desired.
By the closed graph theorem, $T$ is thus a bounded linear operator. Clearly, $T$ is bijective, so that $T^{-1}$ is also a bounded operator, by the bounded inverse theorem. Since $T$ and $T^{-1}$ are bounded, the norms $\|\cdot\|$ and $|\cdot|$ are equivalent, contradicting our assumption that they are not.$\square$
Given a sequence $(x_n)_{n \in \mathbb{N}}$ and $x,y$ as provided by the lemma, we can now construct the map $g$. First, define $$ f_1 : (-1,1) \to F_1, t \mapsto \begin{cases} x_{n+1} + \frac{|t| - (n+1)^{-1}}{n^{-1} - (n+1)^{-1}} \cdot (x_n - x_{n+1}), & \text{if } |t| \in [(n+1)^{-1}, n^{-1}) \text{ for some } n \in \mathbb{N}, \\ x, & \text{if } t = 0 \, . \end{cases} $$ It is not hard to check that $f_1$ is continuous (but in fact, we will not even need this). Furthermore, for $|t| \in [(n+1)^{-1}, n^{-1})$, we have $\gamma := \frac{|t| - (n+1)^{-1}}{n^{-1} - (n+1)^{-1}} \in [0,1]$ and thus $$ \|f_1 (t) - f_1(0)\| = \| \gamma \cdot (x_n - x) + (1-\gamma) \cdot (x_{n+1} - x) \| \leq \max \{ \|x_n - x\| , \|x - x_{n+1}\| \} \, , $$ which goes to zero as $t \to 0$, since then $n \to \infty$. Thus, $g_1$ is continuous at $0$. In exactly the same way, one sees that $$ f_2 : (-1,1) \to F_2 , t \mapsto \begin{cases} x_{n+1} + \frac{|t| - (n+1)^{-1}}{n^{-1} - (n+1)^{-1}} \cdot (x_n - x_{n+1}), & \text{if } |t| \in [(n+1)^{-1}, n^{-1}) \text{ for some } n \in \mathbb{N}, \\ y, & \text{if } t = 0 \end{cases} $$ is continuous at $0$.
Thus, if we define $g_1 : (-1,1) \to F_1, t \mapsto t \cdot f_1 (t)$, then $g_1$ is differentiable at $t = 0$, with $D g_1 (0) = x$, since $$ \lim_{t \to 0} \frac{g_1(t) - g_1 (0)}{t} = \lim_{t \to 0} \frac{t \cdot f_1(t)}{t} = \lim_{t \to 0} f_1 (t) \to f_1 (0) = x \, , $$ with convergence of the limit in $F_1$. In precisely the same way, we see that $g_2 : (-1,1) \to F_2, t \mapsto t \cdot f_2 (t)$ is differentiable at $0$ with $D g_2 (0) = y \neq x$. Finally, note that $g_1 (t) = g_2 (t)$ for all $t \in (-1,1)$; for $t \neq 0$ this is clear and for $t = 0$, we have $g_1 (0) = 0 = g_2 (0)$.$\square$
Remark 1 Above, I committed the usual abuse of notation of identifying an element $x \in F$ of a vector space with the linear map $\mathbb{R} \to F, t \mapsto t \cdot x$.
Remark 2 By slightly modifying the construction (in fact, things get easier), one can show that there is a function $g : (-1,1) \to F$ such that as a map into $F_1$, $g$ is differentiable at $0$, but as a map into $F_2$, it is not.