Monotonicity of $y(x)=1+xy(x)^m$ I want to know the monotonicity of this function $y(x)$ defined by
$$y(x)=1+xy(x)^m,\quad x\in[-\frac{(m-1)^{m-1}}{m^m},\frac{(m-1)^{m-1}}{m^m}]$$
A power expansion of $y(x)$ is $$y(x)=\sum\limits_{n=0}^\infty \binom{mn}{n}\frac{x^n}{(m-1)n+1},\quad y(0)=1,$$
where $m>1$ is a positive integer. 
 A: Incomplete answer:
Consider the inverse:
$$y^{-1}=\frac{x-1}{x^m}=x^{1-m}-x^{-m}$$
The derivative is given by $(1-m)x^{-m}+mx^{-1-m}$, which is equal to $0$ when
$$(m-1)x^{-m}=mx^{-1-m}$$
$$x=\frac m{m-1}$$
Plugging this back in, we get
$$y^{-1}=\frac{(m-1)^{m-1}}{m^{m-1}}-\frac{(m-1)^m}{m^m}$$
There is also the additional point $x=0$, where the derivative of the inverse becomes undefined and approaches $\pm\infty$ as $x\to0$, depending on the parity of $m$. Since it approaches infinity, we can ignore this case.
And so you have a monotone function over any interval not containing $\frac{(m-1)^{m-1}}{m^{m-1}}-\frac{(m-1)^m}{m^m}$, assuming $y$ uniquely exists. It is clear from the series expansion that this is increasing on $\left[0,\frac{(m-1)^m}{m^m}\right]$, and thus increasing on $\left[\frac{(m-1)^{m-1}}{m^{m-1}}-\frac{(m-1)^m}{m^m},\frac{(m-1)^m}{m^m}\right]$, but whether or not this is increasing near $-\frac{(m-1)^m}{m^m}$ I am unsure.
A: Consider the function
$$
 f: \begin{cases} (0,\frac{m}{m-1}] &\to \Bbb R, \\
y &\mapsto \frac{y-1}{y^m} \,.
\end{cases}
$$
On the open interval $(0,\frac{m}{m-1})$ we have
$$
 f'(y)= \frac{m-(m-1)y}{y^{m+1}} > 0
$$
so that $f$ is strictly increasing. Also
$$
 \lim_{y \to 0} f(y) = -\infty \,, \quad f(\frac{m}{m-1}) = \frac{(m-1)^{m-1}}{m^{m}} \, .
$$
It follows that $f$ is a bijective (increasing) mapping from $(0,\frac{m}{m-1}]$ to $(-\infty, \frac{(m-1)^{m-1}}{m^{m}}]$. The inverse function $g = f^{-1}$ is defined on $(-\infty, \frac{(m-1)^{m-1}}{m^{m}}]$ and satisfies 
$$
 f(g(x)) = x \implies g(x) = 1 + x g(x)^m
$$
on that interval. As the inverse function of a (strictly) increasing function, $g$ is strictly increasing as well.
