Not as complicated as it looks

Note that $f(x)=\dfrac{x^{2}(x+1)^{2}}{4}$ , $g(x)=\dfrac{x(x+1)(2x+1)}{6}$ , and $k(x)=\dfrac{x(x+1)}{2}$ .

So,

$\left ( \large \dfrac{\frac{f(x)}{g(x)}}{k(x)} \right )^{-1}=\left ( \large \dfrac{\frac{x^{2}(x+1)^{2}}{4}\cdot \frac{6}{x(x+1)(2x+1)}}{\dfrac{x(x+1)}{2}} \right )^{-1}=\left ( \dfrac{3x(x+1)}{2(2x+1)}\cdot \dfrac{2}{x(x+1)} \right )^{-1}$

$=\left ( \dfrac{3}{2x+1} \right )^{-1}=\dfrac{2x+1}{3}$

So, $\large \left ( \dfrac{\frac{f(1)}{g(1)}}{k(1)} \right )^{-1}+\left ( \dfrac{\frac{f(2)}{g(2)}}{k(2)} \right )^{-1}+\left ( \dfrac{\frac{f(3)}{g(3)}}{k(3)} \right )^{-1}+\cdots +\left ( \dfrac{\frac{f(100)}{g(100)}}{k(100)} \right )^{-1}$ becomes

$\dfrac{3}{3}+\dfrac{3+2}{3}+\dfrac{3+4}{3}+\cdots+\dfrac{3+198}{3}$ , which simplifies to

$\dfrac{300+9900}{3}=\boxed{3400}$ .

$\begin{aligned} S & = \sum_{x=1}^{100} \left(\frac {\frac {f(x)}{g(x)}}{k(x)}\right)^{-1} \\ & = \sum_{x=1}^{100} \left(\frac {\frac{x^2(x+1)^2}{2^2} \cdot \frac 6{x(x+1)(2x+1)}}{\frac {x(x+1)}2}\right)^{-1} \\ & = \sum_{x=1}^{100} \left(\frac{x^2(x+1)^2}{2^2} \cdot \frac 6{x(x+1)(2x+1)} \cdot \frac 2{x(x+1)} \right)^{-1} \\ & = \sum_{x=1}^{100} \left(\frac 3{2x+1} \right)^{-1} \\ & = \sum_{x=1}^{100} \frac {2x+1}3 \\ & = \frac 13 \sum_{x=1}^{100} (2x+1) \\ & = \frac 13 (100(101) + 100) \\ & = \boxed{3400} \end{aligned}$