If $f\left(x\right)+2f(1/x)=3x, x\ne 0$ and $\left\{x\in R:f\left(x\right)=f(-x)\right\}$; then S 

$f\left(x\right)+2\left(\frac{{\displaystyle 1}}{{\displaystyle x}}\right)=3x\to \left(1\right)$

Replacing x by $\frac{1}{x}$ we get $f\left(\frac{{\displaystyle 1}}{{\displaystyle x}}\right)+2f\left(x\right)=\frac{3}{x}\to \left(2\right)$

$2\left(2\right)-\left(1\right)\Rightarrow 3f\left(x\right)=\frac{6}{x}-3x\Rightarrow f\left(x\right)=\frac{2}{x}-x$

$f\left(x\right)=f(-x)\Rightarrow \frac{2}{x}-x=-\frac{2}{x}+x\Rightarrow 2x=\frac{4}{x}\Rightarrow x^2=2\Rightarrow x=\pm \sqrt{2}$

$\therefore$S contains exactly two elements