Let $k\in \mathbb{N}$. I'm interested in for what $k$ the function $\frac{1}{1+\|\cdot\|}f_{k}\in L^1(\mathbb{R}^d)$ and $\partial_{x_{i}}f_{k}\in L^1(\mathbb{R}^d)$
Where $f_k:\mathbb{R}^d\to\mathbb{R}^d$ is defined as
$$ f_k(x)=\frac{x}{\|x\|^k}. $$
If this is too much to ask then a simpler thing to ask ( which Im still interested in) is for what $k$ is $f_k\in L^1(\mathbb{R}^d)$?
$\textbf{Note : }$ (since the function looks radial) I tried to use polar coordinates (assuming $d$ is even) : $(x_i,x_j)=r_i(\cos \theta_i,\sin \theta_i)$ with $r_i\in \mathbb{R}_+,\theta_i\in [0,2\pi)$, so that $$dx=\prod_i r_idrd\theta$$
and
$$ \int_{\mathbb{R}_+^{d/2} \times [0,2\pi]^{d/2}}\big|\big(f(r,\theta)\big)_i\big|\prod_i r_idrd\theta =\int_{\mathbb{R}^d} \frac{| r_i \cos \theta_i |}{\|r\|^{k}} \prod_i r_idrd\theta $$
then I was thinking of using that $\Big(\prod_i r_i\Big)^{2/d}\leq \frac{2}{d}\sum_i^{d/2} r_i$ but im really not sure where im going with this.
This is not how polar coordinates work in $\mathbb{R}^d$. Their construction is rather difficult and german wikipedia should give you a nice impression. You could also use the coarea formula that states $$ \int_{\mathbb{R}^d}u(x)~\mathrm{d}x = \int^\infty_0\int_{\partial B_r(0)} u(x)~\mathrm{d}\sigma(x)~\mathrm{d}r $$ for $u \in L^1(\mathbb{R}^d)$.
The important takeaway from this should be that there exists some constant $C >0 $ (that is some integral over a finite product of $\sin$ and $\cos$ that you get in the numerator when you pass to polar coordinates) such that: $$ \int_{\mathbb{R}^d} \frac{\lvert x_j \rvert}{(1+\lVert x \rVert)\lVert x \rVert^k}~\mathrm{d}x = C\int^\infty_0 \frac{r}{(1+r)r^k} \underbrace{r^{d-1}}_{\text{Jacobian}}~\mathrm{d}r = C\int^\infty_0 \frac{r^{d-k}}{1+r}~\mathrm{d}r $$ This does never converge.
Nothing stops you from applying this reasoning to the derivative, too.