The function $f$ is not defined for $x=0$, but, for all non-zero real numbers $x$, $f(x)+2f\left(\dfrac{1}x\right)=3x$. The equation $f(x)=f(-x)$ is satisfied by
$\textbf{(A)}\ \text{exactly one real number} \qquad \textbf{(B)}\ \text{exactly two real numbers} \qquad\textbf{(C)}\ \text{no real numbers}\qquad \\ \textbf{(D)}\ \text{infinitely many, but not all, non-zero real numbers} \qquad\textbf{(E)}\ \text{all non-zero real numbers}$
Substitute $x$ with $\frac{1}{x}$:
$f(\frac{1}{x})+2f(x)=\frac{3}{x}$.
Adding this to $f(x)+2f\left(\dfrac{1}x\right)=3x$, we get
$3f(x)+3f\left(\dfrac{1}x\right)=3x+\frac{3}{x}$, or
$f(x)+f\left(\dfrac{1}x\right)=x+\frac{1}{x}$.
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 $x=\pm{\sqrt2}$ are the two real solutions and the answer is $\boxed{(B)}$.