There are two equations showed in Gel'fand and Shilov's book (Generalized Functions I Properties and Operations) on page 183 and 185:
$${\delta}^{(k-1)}(1-x^2)=\frac{(-1)^{k-1}}{2^kx^{k-1}}[{\delta}^{(k-1)}(x-1)-{\delta}^{(k-1)}(x+1)];\tag{1}$$
$${\delta}^{(k)}(f(x))=\sum_n \frac{1}{|f'(x_n)|}(\frac{1}{f'(x)}\frac{d}{dx})^k{\delta}(x-x_n),\tag{2}$$ where $x_n$ are $n$ simple roots of $f(x).$
I think the first equation should satisfy the second equation. But it seems not. Why? And I would like to know the right equation.
TL;DR: Formula (2) is correct, which can easily be checked$^1$ by using test functions. However, formula (1) is incorrect.
We calculate$^2$ $$ u_k(x)~:=~(-1)^k\delta^{(k)}(1\!-\!x^2) ~=~\delta^{(k)}(x^2\!-\!1)~=~\left.\delta^{(k)}(y) \right|_{y=x^2-1}$$ $$~=~\left.\left(\frac{d}{dy}\right)^k\delta(y)\right|_{y=x^2-1} ~=~\left(\frac{1}{2x}\frac{d}{dx}\right)^k\delta(x^2\!-\!1) ~=~\left(\frac{1}{2x}\frac{d}{dx}\right)^k\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1). \tag{A}$$
By anti-normal-ordering (=ordering derivatives to the left) of eq. (A) we get for the first few $k\in\mathbb{N}_0$:
$$\begin{align}u_{k=0}(x)&~=~\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1),\cr u_{k=1}(x)&~=~\frac{1}{2x}\frac{d}{dx}\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1)\cr &~=~\left(\frac{d}{dx}\frac{1}{2x}+\frac{1}{2x^2}\right)\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1)\cr &~=~\frac{1}{4}\sum_{\pm}\pm \delta^{\prime}(x\!\mp\!1)+\frac{1}{4}\sum_{\pm}\delta(x\!\mp\!1),\cr u_{k=2}(x)&~=~\left(\frac{1}{2x}\frac{d}{dx}\right)^2\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1)\cr &~=~\left(\frac{d^2}{dx^2}\frac{1}{4x^2}+\frac{d}{dx}\frac{3}{4x^3}+\frac{3}{4x^4}\right)\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1)\cr &~=~\frac{1}{8}\sum_{\pm}\delta^{\prime\prime}(x\!\mp\!1)+\frac{3}{8}\sum_{\pm}\pm \delta^{\prime}(x\!\mp\!1)+\frac{3}{8}\sum_{\pm}\delta(x\!\mp\!1),\cr &~~~\vdots\end{align}\tag{B}$$
By normal-ordering (=ordering derivatives to the right) of eq. (A) we get for the first few $k\in\mathbb{N}_0$:
$$\begin{align}u_{k=0}(x)&~=~\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1),\cr u_{k=1}(x)&~=~\frac{1}{2x}\frac{d}{dx}\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1)\cr &~=~\frac{1}{4x}\sum_{\pm}\delta^{\prime}(x\!\mp\!1),\cr u_{k=2}(x)&~=~\left(\frac{1}{2x}\frac{d}{dx}\right)^2\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1)\cr &~=~\left(\frac{1}{4x^2}\frac{d^2}{dx^2}-\frac{1}{4x^3}\frac{d}{dx}\right)\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1)\cr &~=~\frac{1}{8x^2}\sum_{\pm}\delta^{\prime\prime}(x\!\mp\!1) -\frac{1}{8x^3}\sum_{\pm} \delta^{\prime}(x\!\mp\!1),\cr &~~~\vdots\end{align}\tag{C}$$ which is different from eq. (1). It is interesting how anti-normal-ordering (B) and normal-ordering (C) generate very different-looking expressions for the same underlying mathematical distributions.
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$^1$ In this answer, we make repeated use of the following distribution identities: $$\delta(f(x))~=~\sum_{i}^{f(x_i)=0}\frac{1}{|f'(x_i)|}\delta(x\!-\!x_i), \tag{D}$$ $$ \{f(x)-f(y)\}\delta(x\!-\!y)~=~0, \tag{E}$$ and derivatives thereof: $$ \left(\frac{d}{dx}\right)^k[\{f(x)-f(y)\}\delta(x\!-\!y)]~=~0. \tag{F}$$
$^2$ We have for convenience shifted OP's definition $k-1\to k$ and removed an over-all sign factor $(-1)^k$. This can of course easily be reinstalled.