Let $E$ be a $\mathbb R$-Banach space, $v:E\to(0,\infty)$ be continuous with $$\inf_{x\in E}v(x)>0\tag1,$$ $r\in(0,1]$ and$^1$ $$\rho(x,y):=\inf_{\substack{c\:\in\:C^1([0,\:1],\:E)\\ c(0)=x\\ c(1)=y}}\int_0^1v^r\left(c(t)\right)\left\|c'(t)\right\|_E\:{\rm d}t\;\;\;\text{for }x,y\in E.$$ Note that $\rho$ is a well-defined metric on $E$. Let $$|f|_{\operatorname{Lip}(\rho)}:=\sup_{\substack{x,\:y\:\in\:E\\x\:\ne\:y}}\frac{|f(x)-f(y)|}{\rho(x,y)}\;\;\;\text{for }f:E\to\mathbb R$$ and $$\operatorname{Lip}(\rho):=\left\{f:E\to\mathbb R\mid|f|_{\operatorname{Lip}(\rho)}<\infty\right\}.$$ $|\;\cdot\;|_{\operatorname{Lip}(\rho)}$ is a semi-norm on $\operatorname{Lip}(\rho)$. Let $\mu$ be a probability measure on $(E,\mathcal B(E))$ with $$\int\rho(\;\cdot\;,0)\:{\rm d}\mu<\infty\tag2$$ By $(2)$, $$\operatorname{Lip}(\rho)\subseteq\mathcal L^1(\mu)$$ and $$\left\|f\right\|_{\operatorname{Lip}(\rho)}:=\left|\int f\:{\rm d}\mu\right|+|f|_{\operatorname{Lip}(\rho)}\;\;\;\text{for }f\in\operatorname{Lip}(\rho)$$ is a norm.

Let $f\in\operatorname{Lip}(\rho)$ be Fréchet differentiable. How can we show that $$\left\|f\right\|_{\operatorname{Lip}(\rho)}=\sup_{x\in E}\frac{\left\|{\rm D}f(x)\right\|_{E'}}{v^r(x)}+\int f\:{\rm d}\mu?\tag3$$ In particular, how can we show that $$\sup_{\substack{y\:\in\:E\\0\:<\left\|x-y\right\|_E\:<\:\varepsilon}}\frac{|f(x)-f(y)|}{\rho(x,y)}\xrightarrow{\varepsilon\to0}\frac{\left\|{\rm D}f(x)\right\|_{E'}}{v^r(x)}\tag4?$$

EDIT: It's clear to me that $$\sup_{y\in B_\delta(x)\setminus\{x\}}\frac{|f(x)-f(y)|}{\left\|x-y\right\|_E}\xrightarrow{\delta\to0+}\left\|{\rm D}f(x)\right\|_{E'}\tag5.$$ So, writing $$\frac{|f(x)-f(y)|}{\rho(x,y)}=\frac{|f(x)-f(y)}{\left\|x-y\right\|_E}\frac{\left\|x-y\right\|_E}{\rho(x,y)}\tag6,$$ we should only need to deal with $\frac{\left\|x-y\right\|_E}{\rho(x,y)}$.