Range of $f(x)= \frac{\tan{x}}{\tan{3x}} $ 
Prove that for the values of $x$ where the following $f(x)$ is defined, $f(x)$ does not lie between $\frac{1}{3}$ and $3$. $$f(x)=\frac{\tan{x}}{\tan{3x}}$$

My Attempt:
I wrote down, $$\tan{3x}=\frac{3\tan{x}-\tan^3{x}}{1-3\tan^2{x}}$$
This reduced $f(x)$ to,
$$f(x)=\frac{1-3\tan^2{x}}{3-\tan^2{x}}$$
I don't know how to solve any further. I thought of using derivative, but the function is dicontinuous at times. How do I solve it? Any hints would be helpful.
Thanks.

Note that $\tan(x)^2 \ge 0$ and 
$$f(x)= \frac{1-3\tan(x)^2}{3-\tan(x)^2} = 3+\frac{8}{\tan(x)^2-3}.$$
Clearly, for $\tan(x)^2 > 3$, $\frac{8}{\tan(x)^2-3} > 0$, and $f(x) >3$.
For $0 \le \tan(x)^2 < 3$, we have $-3 \le \tan(x)^2-3 < 0 \implies \frac{8}{\tan(x)^2-3}\le-\frac{8}{3} \implies f(x) = 3+\frac{8}{\tan(x)^2-3} \le 3-\frac{8}{3} = \frac{1}{3}.$
