It is given function \(f(x)\) for all real number \(x\) such that: \(x(2x+1)f(x)+f(\frac{1}{x})=x+1\).

From \(x(2x+1)f(x)+f(\frac{1}{x})=x+1\), replacing \(x\) by \(\frac{1}{x}\) gives \(f(\frac{1}{x})+\frac{x^2}{x+2}f(x)=\frac{x^2+x}{x+2}\). Subtracting, \(\frac{2x(x^2+1)}{x+2}f(x)=\frac{2x+2}{x+2}\), hence \(f(x)=\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}\). Therefore, \(S=\sum_{k=1}^{2022}f(k)=\sum_{k=1}^{2022}(\frac{1}{k}-\frac{1}{k+1})=1-\frac{1}{2022}=\frac{2021}{2022}\).