I am looking for an explicit cut-off function $\xi(x)\in C^\infty_c(\mathbb{R}^d)$ such that $|\nabla_x\xi(x)|\leq C\xi(x)$ for some $C>0$. What I have found is the following one. Consider the two balls $B_m$ and $B_{m-1}$ of radius $m$ and $m-1$ both respectively centred in $0$. We can define $\xi_m(x)$ as $1$ in $B_{m-1}$, $0$ outside $B_m$, and $$\xi_m(x)=\frac{\exp\left\{\frac{1}{|x|_d-m}\right\}}{\exp\left\{\frac{1}{|x|_d-m}\right\}+\exp\left\{-\frac{1}{|x|_d-(m-1)}\right\}}\quad\text{ for }x\in B_m \setminus B_{m-1}. $$
Now, if I consider the gradient I am able to prove only that $$|\nabla_x\xi_m(x)|^2 \leq C \xi(x)$$ and it fails the property that I am looking for ($|\nabla_x\xi_m(x)| \leq C \xi(x)$).
Do you have any suggestions?
No, this is not possible. Let's go to $d=1$ for a counterexample. Assume $f\in C^1(\mathbb R ),$ with $f> 0$ on $(0,1)$ and $f=0$ elsewhere. Suppose, to reach a contradiction, that
$$|f'(x)|\le C|f(x)|$$
for $x\in (0,1).$ For such $x$ we then have
$$\tag 1 |f(x)|=|\int_0^x f'| \le \int_0^x |f'|\le C\int_0^x |f|.$$
For each such $x,$ define $M_x=\max_{[0,x]} |f|.$ By continuity, there exists $y(x)\in [0,x]$ such that $M_x= |f(y(x))|.$ So from $(1)$ we see
$$\tag 2 M_x = |f(y(x))| \le C\int_0^{y(x)} |f| \le C\int_0^{x} |f|\le CxM_x.$$
Now $M_x>0,$ so we can divide by $M_x$ in $(2)$ to get $1\le Cx.$ This is true for all $x\in (0,1).$ Clearly this is impossible.