= Solution
By <equivalent norms>, choose $c,C>0$ with $c\|h\|_1\le\|h\|_2\le C\|h\|_1$. Differentiability in the first <norm> means that there is a bounded <linear map> $L:V\to\mathbb R$ with $f(a+h)-f(a)-Lh=r(h)$ and $|r(h)|/\|h\|_1\to0$. The same $L$ is bounded in the second <norm>, since $|Lh|\le K\|h\|_1\le Kc^{-1}\|h\|_2$. Moreover,
$$
\frac{|r(h)|}{\|h\|_2}\le\frac1c\frac{|r(h)|}{\|h\|_1}\longrightarrow0.
$$
The two <norms> define the same approach $h\to0$, so this proves differentiability with the same <Fréchet derivative> in the second <norm>. Reversing the <norms> proves the converse.
For completeness, the <derivative> is unique: if $L_1,L_2$ both work, fix $v$ and take $h=tv$. The two remainder estimates imply $|(L_1-L_2)v|=o(1)$ as $t\to0$, so $(L_1-L_2)v=0$ for every $v$. Thus \b[differentiability and its <derivative> are independent of the chosen <norm>]. This is <differentiability under equivalent norms>, not merely invariance of continuity.
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