Consider the function $\Vert\cdot\Vert_{L^2}:H^1(\mathbb{R})\to \mathbb{R}$ and a point $0\neq f\in H^1(\mathbb{R})$. Is $\Vert\cdot\Vert_{L^2}$ Fréchet differentiable w.r.t. to the norm $\Vert\cdot\Vert_{H^1}$? I.e. is there a bounded linear operator $A:H^1(\mathbb{R})\to \mathbb{R}$, s.t.
\begin{equation} \lim_{\Vert g\Vert_{H^1}\to 0} \frac{\left\lvert \Vert f+g\Vert_{L^2} - \Vert f\Vert_{L^2} - Ag\right\rvert}{\Vert g\Vert_{H^1}} = 0? \end{equation}
Yes. We may show (see here for example) that, for a Hilbert space $H$, the derivative of the norm $\|\cdot\|$ at $f \in H$ is given by $$A(f)(g) = \left\langle g, \frac{f}{\|f\|}\right\rangle.$$
This implies that, in the $L^2$ case, \begin{equation} \lim_{\Vert g\Vert_{L^2}\to 0} \frac{\left\lvert \Vert f+g\Vert_{L^2} - \Vert f\Vert_{L^2} - A(f)(g)\right\rvert}{\Vert g\Vert_{L^2}} = 0. \end{equation} As $\|g\|_{L^2} \le \|g\|_{H^1}$, we have $$\frac{\left\lvert \Vert f+g\Vert_{L^2} - \Vert f\Vert_{L^2} - A(f)(g)\right\rvert}{\Vert g\Vert_{H^1}} \le \frac{\left\lvert \Vert f+g\Vert_{L^2} - \Vert f\Vert_{L^2} - A(f)(g)\right\rvert}{\Vert g\Vert_{L^2}}$$ and we conclude the desired result.