smc

Substitute $x$ with $\frac{1}{x}$: $f(\frac{1}{x})+2f(x)=\frac{3}{x}$. Adding this to $f(x)+2f\left(\frac{1}{x}\right)=3x$, we get $3f(x)+3f\left(\frac{1}{x}\right)=3x+\frac{3}{x}$, or $f(x)+f\left(\frac{1}{x}\right)=x+\frac{1}{x}$. Subtracting this from $f(\frac{1}{x})+2f(x)=\frac{3}{x}$, we have $f(x)=\frac{2}{x}-x$. Then, $f(x)=f(-x)$ when $\frac{2}{x}-x=\frac{2}{-x}+x$ or $\frac{2}{x}-x=0$, so $x=\pm \sqrt{2}$ are the two real solutions and the answer is $\boxed{\text{exactly two real numbers}}$.