Calculate, $f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots f\bigg(\frac{1996}{1997}\bigg)$

I would say your method is practically speaking what I would also do. Maybe I would rephrase it as follows:

Claim: $f(a)+f(1-a)=1$.

Then write $S$ for the sum in question, and then $2S$ can be written as $f(1/1997+1996/1997) + \cdots$ (the Gauss trick), which is $1996$ by the claim, so $S=998$.

Given $f(x)=\dfrac{4^x}{4^x+2}$

From that we get $f(1-x)=\dfrac{2}{4^x+2}$

First let us take the last term $f\left(\dfrac{1}{1997}\right)$

Notice that $f\left(\dfrac{1}{1997}\right)=f\left(1-\dfrac{1}{1997}\right)$ and same for the rest of the terms.

Now, $f(x)+f(1-x)=1$

$f\left(\dfrac{1}{1997}\right)+f\left(\dfrac{2}{1997}\right)+........+f\left(1-\dfrac{2}{1997}+f\left(1-\dfrac{1}{1997}\right)\right)$

all of them makes pairs.

So, the total pairs $=\dfrac{1996}{2}=998$

So, the sum $=1+1+1+......998$ times $=998$