Find $k$ in Maclaurin series expansion of $\frac{dy}{dx}=-\frac{1}{2}+\frac{1}{4}x+kx^2+...$ where $y=\ln\Bigl(\frac{e^{-x}+1}{2}\Bigl)$ 
Given that $y=\ln\Bigl(\frac{e^{-x}+1}{2}\Bigl)$, show $\frac{dy}{dx}=\frac{1}{2}e^{-y}-1$. Show that the series expansion of $\frac{dy}{dx}$ in ascending powers of $x$, up to and including the term in $x^2$ is $-\frac{1}{2}+\frac{1}{4}x+kx^2+\dots$ , where $k$ is to be determined.

I'm able to solve this question but I'm unsure if my $k$ value is correct. My $k$ is $0$. Am I correct? Otherwise, I might have made a mistake somewhere.
My work
1) $f(0)=y=0$
2) $\frac{dy}{dx}=\frac{1}{2}e^{-y}-1\ $
therefore $f'(0)=-\frac{1}{2}$
3) $\frac{d^2y}{dx^2}=-\frac{1}{2}e^{-y}\cdot\Big(\frac{dy}{dx}\Bigl)\ $
therefore $f''(0)=\frac{1}{4}$
4) $\frac{d^3y}{dx^3}=\frac{1}{2}e^{-y}\cdot\Bigl(\frac{dy}{dx}\Bigl)^2-\frac{1}{2}e^{-y}\cdot\Bigl(\frac{d^2y}{dx^2}\Bigl)\ $
therefore $f'''(0)=0$
Hence
$$\frac{dy}{dx}=f'(0)+f''(0)x+\frac{f'''(x)}{2}x^2+\dots=-\frac{1}{2}+\frac{1}{4}x+\dots$$

Yes, you are correct $k=0$. This is an alternative solution  where we use the expansions of $e^t$ and $(1+t)^{-1}$ at $t=0$:
$$\begin{align}
\frac{dy}{dx}&=\frac{2}{e^{-x}+1}\cdot \frac{-e^{-x}}{2} =-\frac{1}{1+e^x}\\
&=-\frac{1}{1+1+x+\frac{x^2}{2}+o(x^2)}\\
&=-\frac{1}{2}\left(1+\frac{x}{2}+\frac{x^2}{4}+o(x^2)\right)^{-1}\\
&=-\frac{1}{2}\left(1-\left(\frac{x}{2}+\frac{x^2}{4}\right)+\left(\frac{x}{2}+o(x)\right)^2+o(x^2)\right)\\
&=-\frac{1}{2}+\frac{x}{4}+\underbrace{\left(-\frac{1}{4}+\frac{1}{4}\right)}_{=0}\cdot x^2+ o(x^2).
\end{align}$$
