In $\mathbb{R}^n$, let
$$w(x)=\sqrt{\sin^2 x_1+\cdots+\sin^2 x_n}\,,$$
and
$$|x|=\sqrt{x_1^2+\cdots+x_n^2}\;.$$
Let
$$f(x)=\frac{1}{w(x)}-\frac{1}{|x|}.$$
Then how to prove that
$$|\partial^{\alpha} f(x)|\leq C_\alpha |x|^{1-|\alpha|},\;\;\mbox{as}\;\;|x|\rightarrow 0,$$
for all multi-index $\alpha\in \mathbb{N}^n$, $C_\alpha$ is a constant depend on $\alpha$.
If $n=1$, then (consider $x>0$ is enough) $$f(x)=\frac{1}{\sin x}-\frac{1}{x}$$ and this case is easy. In higher dimensions, I want to expand $f(x)$ at $0$ but it is a little hard.
Comments: We can reduce this question to estimate $$\partial^{\alpha}(|x|f(x))=\partial^{\alpha}\left(\frac{|x|}{w(x)}\right)$$ for $|\alpha|\geq 1$. We eliminate the radical and consider the derivative of $$g(x):=\frac{|x|^2}{w^2(x)}.$$ Then it suffices to consider $$\partial^{\alpha}(w^2(x)g(x))=\partial^{\alpha}(|x|^2)$$ and by an induction we can derive the desired conclusion.