How to prove $\exp(x)/(\exp(x)+1)^2$ is even?

Hint:

$$\frac{e^x}{(e^x+1)^2} = \frac{e^x}{e^{2x}+2e^x+1}.$$

Now try multiplying the numerator and denominator by a certain term to get something more symmetric looking. An expanded hint below in the spoiler text if you want to try to figure it out for yourself from here without getting the full answer.

Multiply and divide by $e^{-x}$.

Start with $$ f(-x) = \frac{\exp(-x)}{(\exp(-x)+1)^2}. $$ Multiplying the numerator by $\exp(2x)$ yields $\exp x$.

Multiplying the denominator by $\exp(2x) = (\exp x)^2$ yields $$ \Big( (\exp x) \, (\exp(-x)+1)) \Big)^2 = \Big( 1+ \exp x\Big)^2. $$

Why are you trying expanding into series? Just use the definition – you need to show $f(-x) = f(x)$, so:

$$ \begin{align} f(-x) & = \frac{e^{-x}}{(e^{-x}+1)^2} \cdot \frac{e^{2x}}{e^{2x}} \\ & = \frac{e^{-x}\cdot e^{2x}}{(e^{-x}+1)^2\cdot (e^x)^2} \\ & = \frac{e^x}{((e^{-x}+1)\cdot e^x)^2} \\ & = \frac{e^x}{(1 + e^x)^2} \\ & = f(x) \end{align} $$

Done.

You could also reduce the expression to a symmetric form $$ f(x) = \frac{e^x}{(e^x+1)^2} \cdot \frac{e^{-x}}{e^{-x}} = \frac 1{(e^x+1)^2\cdot (e^{-x/2})^2} = \frac 1{(e^{x/2}+e^{-x/2})^2} $$ and conclude both terms in the denominator replace each other with the $x$ sign switch, thus preserving the value of a sum, hence $f$ is even.