We have, $f(x)+2 f\left(\frac{1}{x}\right)=3 x, \quad x \neq 0$   $\ldots$ (i)

On replacing $x$ by $\frac{1}{x}$ in the above equation, we get

$$ \begin{aligned} & f\left(\frac{1}{x}\right)+2 f(x) =\frac{3}{x} \\\\ \Rightarrow & \,\, 2 f(x)+f\left(\frac{1}{x}\right) =\frac{3}{x} \,\,\,\,\,...(ii) \end{aligned} $$

On multiplying Eq. (ii) by 2, we get

$$ 4 f(x)+2 f\left(\frac{1}{x}\right)=\frac{6}{x}\quad...(iii) $$

and subtracting Eq. (i) from Eq. (iii), we get

$$[4 f(x)+2 f\left(\frac{1}{x}\right)] - [f(x)+2 f\left(\frac{1}{x}\right)]=\frac{6}{x} - 3x$$

$\Rightarrow {3 f(x)=\frac{6}{x}-3 x}$

$\Rightarrow f(x)=\frac{2}{x}-x$

Now, consider $\quad f(x)=f(-x)$

$$ \begin{aligned} &\Rightarrow \frac{2}{x}-x =-\frac{2}{x}+x \\\\ &\Rightarrow \frac{4}{x} =2 x \\\\ &\Rightarrow 2 x^2 =4 \\\\ &\Rightarrow x^2 =2 \\\\ &\Rightarrow x =\pm \sqrt{2} \end{aligned} $$

Hence, $S$ contains exactly two elements.