Let $f(x)=\arcsin(1-x)$ for $x\in[0,2]$.
Since the derivative of $f(x)=O\left( x^{-1/2}\right)$ for $x\sim 0$, we let $t=x^{1/2}$ and $g(t)=\arcsin(1-t^2)$.
We will now develop the first few terms of the Taylor series for $g(t)$ around $t=0$.
We have for the first derivative $g^{(1)}(t)$
$$\begin{align}
g^{(1)}(t)&=-\frac{2t}{\sqrt{1-(1-t^2)^2}}\\\\
&=-\frac{2}{\sqrt{2-t^2}}\tag 1
\end{align}$$
Differentiating the right-hand side of $(1)$, we obtain the second derivative, $g^{(2)}(t)$
$$\begin{align}
g^{(2)}(t)&=-\frac{2t}{(2-t^2)^{3/2}}\tag 2
\end{align}$$
Continuing, we have for $g^{(3)}(t)$
$$\begin{align}
g^{(3)}(t)&=-\frac{4(t^2+1)}{(2-t^2)^{5/2}}\tag 3
\end{align}$$
And finally, we have for $g^{(4)}(t)$
$$\begin{align}
g^{(4)}(t)&=-\frac{12t(t^2+3)}{(2-t^2)^{7/2}}\tag 4
\end{align}$$
We evaluate $(1)-(4)$ at $t=0$ and form the expansion
$$\bbox[5px,border:2px solid #C0A000]{\arcsin(1-x)=\frac{\pi}{2}-\sqrt{2}x^{1/2}-\frac{\sqrt{2}}{12}x^{3/2}+O\left(x^{5/2}\right)}$$
Look at the Boustrophedon table:
$$\matrix{1\\0&1\\1&1&0\\0&1&2&2\\5&5&4&2&0\\0&5&10&14&16&16\\61&61&56&46&32&16&0\\0&61&122&178&224&256&272&272\\1385&1385&1324&1202&1024&800&544&272&0}$$
etc. Each row is the series of partial sums of the previous row, but
at each stage one reverses the order we add up and enter the partial sums.
Any, from the first column we read off
$$\sec x=1+\frac{x^2}{2!}+\frac{5x^4}{4!}+\frac{61x^6}{6!}+\frac{1385x^8}{8!}+\cdots.$$
The right-most elements also give
$$\tan x=x+\frac{2x^3}{3!}+\frac{16x^5}{5!}+\frac{272x^7}{7!}+\cdots.$$
There's a good discussion on this in Concrete Mathematics by Graham, Knuth and Patashnik.
Best Answer
There seem to be many ways to go about this, so here is one: put $\dfrac{x}{x ^ 2 + x + 1}=\sum_{n=0}^{\infty} a_nx^n$, then $$x=\sum_{n=0}^{\infty} a_nx^n(x^2+x+1)=a_0+(a_1+a_0)x+\sum_{n=2}^{\infty}(a_n+a_{n-1}+a_{n-2})x^n,$$ and by comparing the coefficients we get $a_0=0$, $a_1=1$, and $a_n+a_{n-1}+a_{n-2}=0$ for $n \geq 2$. You can see the coefficients repeat ($a_2=-1,a_3=0,a_4=1,\dots$), so we have $a_{3k}=0, a_{3k+1}=1, a_{3k+2}=-1$, or in other words $$ \dfrac{x}{x ^ 2 + x + 1}=x-x^2+x^4-x^5+\dots $$