1. Given functional equation

We have $f(x)+2f\left(\frac1x\right)=3x, \qquad x\neq 0.$

We need the set $S=\{x\in \mathbb R: f(x)=f(-x)\}.$

Since the equation is defined only for $x\neq 0$, we work on $\mathbb R\setminus\{0\}$.

2. Find an explicit formula for $f(x)$

Write the given equation for $x$: $f(x)+2f\left(\frac1x\right)=3x \qquad ...(1)$

Now replace $x$ by $\frac1x$: $f\left(\frac1x\right)+2f(x)=\frac{3}{x} \qquad ...(2)$

Let $a=f(x), \qquad b=f\left(\frac1x\right).$ Then (1) and (2) become $a+2b=3x,$ $2a+b=\frac{3}{x}.$

Solve this system.

From the first equation, $a=3x-2b.$ Substitute into the second: $2(3x-2b)+b=\frac{3}{x}$ $6x-4b+b=\frac{3}{x}$ $6x-3b=\frac{3}{x}$ $2x-b=\frac{1}{x}$ $b=2x-\frac{1}{x}.$

Hence $a=3x-2\left(2x-\frac{1}{x}\right)=3x-4x+\frac{2}{x}=\frac{2}{x}-x.$

So, $f(x)=\frac{2}{x}-x, \qquad x\neq 0.$

3. Compute $f(-x)$

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

We want $f(x)=f(-x).$ So, $\frac{2}{x}-x=x-\frac{2}{x}.$

Bring all terms to one side: $\frac{2}{x}+\frac{2}{x}-x-x=0$ $\frac{4}{x}-2x=0.$

Multiply by $x$ (valid since $x\neq 0$): $4-2x^2=0$ $x^2=2.$

Thus, $x=\pm \sqrt{2}.$

4. Determine the set $S$

Therefore, $S=\{-\sqrt2,\sqrt2\}.$ So $S$ contains exactly two elements.

5. Check options

• A: empty set — false
• B: exactly one element — false
• C: exactly two elements — true
• D: more than two elements — false

Hence the correct option is $\boxed{\text{C}}$

6. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.