Not very tough

Consider $f(x)+f(1-x)$ :

$\begin{aligned} f(x) + f(1-x) & = \frac {9^x}{9^x+3} + \frac {9^{1-x}}{9^{1-x}+3} \\ & = \frac {9^x}{9^x+3} + \frac {9^x\cdot 9^{1-x}}{9^x(9^{1-x}+3)} \\ & = \frac {9^x}{9^x+3} + \frac 9{9+3\cdot9^x} \\ & = \frac {9^x}{9^x+3} + \frac 3{3+9^x} \\ & = 1 \end{aligned}$

Now, we have:

$\begin{aligned} S & = \small f \left(\frac{1}{2018}\right) + f\left(\frac{2}{2018}\right) +f\left(\frac{3}{2018}\right)+\cdots +f\left(\frac{2015}{2018}\right) +f\left(\frac{2016}{2018}\right) +f\left(\frac{2017}{2018}\right) \\ & = \small f \left(\frac{1}{2018}\right) + f\left(\frac{2017}{2018}\right) +f\left(\frac{2}{2018}\right)+f\left(\frac{2016}{2018}\right)+\cdots +f\left(\frac{1008}{2018}\right) +f\left(\frac{1010}{2018}\right) +f\left(\frac{1009}{2018}\right) \\ & = \small \underbrace{f \left(\frac{1}{2018}\right) + f\left(1-\frac{1}{2018}\right)}_{= \ 1} +\underbrace{f\left(\frac{2}{2018}\right)+f\left(1-\frac{2}{2018}\right)}_{= \ 1}+\cdots +\underbrace{f\left(\frac{1008}{2018}\right) +f\left(1-\frac{1008}{2018}\right)}_{= \ 1} +\frac 12 \underbrace{\left(f\left(\frac{1009}{2018}\right) + f\left(1-\frac{1009}{2018}\right) \right)}_{= \ 1} \\ & = \underbrace{1+1+1+\cdots+1} _{\text{Number of 1's } = 1008} + \frac 12 \\ & = \boxed{1008.5} \end{aligned}$

Don't know what it has to do with the given function, but anyhow:

$1+2+3+ \dots + 2016+2017 = \frac{1}{2}\cdot 2017\cdot 2018$ .

Since the de numerator is 2018, the result is 1008.5.