Expanding x/(eˣ − 1)

Identity. $\left(1+ax+bx^2\right)\left(1+\frac x2+\frac{x^2}6\right)=1+o(x^2)$.

Expanding. The coefficient of $x$ is $a+\frac12$ and the coefficient of $x^2$ is $b+\frac a2+\frac16$. Both must vanish:
$$a=-\frac12,\qquad b=-\frac a2-\frac16=\frac14-\frac16=\frac1{12}.$$

Result.
$$\frac x{e^x-1}=1-\frac x2+\frac{x^2}{12}+o(x^2).$$

Remark. The coefficients $1,-\frac12,\frac16,0,-\frac1{30},\dots$ (after multiplying by $n!$) are the Bernoulli numbers. Here $\frac{x^2}{12}=\frac{B_2}{2!}x^2$ with $B_2=\frac16$. Since $\varphi(x)+\frac x2$ is even, all the odd coefficients after the first vanish.