To get started: If $c:[0,1]\to E$ is a $C^1$-curve from $x$ to $y$ you have $$ |f(x)-f(y)|=\left|\int_0^1(f\circ c)'(t)dt\right|\le \int_0^1 \|Df(c(t))\|_{E'} \|c'(t)\|dt $$ $$=\int_0^1 \|Df(c(t))\|_{E'} \frac{v^r(c(t))}{v^r(c(t))} \|c'(t)\|dt$$ $$ \le \sup\{\|Df(z)\|_{E'}/v^r(z):z\in E\} \varrho(x,y)(1+\varepsilon)$$ for a suitable curve.
This gives you an inequality for (4) (I guess that $x$ is fixed, there). For the other inequality I would try to take a direction $r\in E$ which (almost) maximizes $|Df(x)(r)|$ and take $y=x+\varepsilon r$. However, I did not try to work this out.