jmc

We are given $$f(x)+2f\left(\frac{1}{x}\right)=3x.$$Replacing $x$ with $\frac{1}{x},$ we get $$f\left(\frac{1}{x}\right)+2f(x)=\frac{3}{x}.$$We can view these equations as a system in $f(x)$ and $f\left(\frac{1}{x}\right).$ Solving for $f(x),$ we find $$f(x)=\frac{2-x^2}{x}.$$Then the equation $f(x)=f(-x)$ becomes $$\frac{2-x^2}{x}=\frac{2-x^2}{-x}.$$Then $2-x^2=x^2-2,$ so $x^2=2.$ The solutions are $\boxed{\sqrt{2},-\sqrt{2}}.$