solve $\tan{x} = \tan{3x}$ I'm asked to solve $\tan{x} = \tan{3x}$
Here's my attempt: 
$$\tan{x} = \tan{3x}$$
$$\tan{x} = \tan{(x + 2x)}$$
$$\tan{x} = \frac{\tan{x} + \tan{2x}}{1-\tan{x}\tan{2x}}$$
Recall the identity:
$$\tan{2x} = \frac{2\tan{x}}{1-\tan^2{x}}$$
So then we have:
$$\tan{x} = \frac{\tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}}{1-\tan{x}\frac{2\tan{x}}{1-\tan^2{x}}}$$
$$\tan{x} - \tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}$$
$$-\tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \frac{2\tan{x}}{1-\tan^2{x}}$$
$$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} \cdot \frac{1-\tan^2{x}}{2\tan{x}} = 1$$
$$\tan^2{x} = -1$$
This does obviously not compute. Why is my way wrong and how can I go about solving it?

When you have this:
$$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \frac{2\tan{x}}{1-\tan^2{x}}$$
It it incorrect to have this next:
$$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} \cdot \frac{1-\tan^2{x}}{2\tan{x}} = 1$$
Just like when you have (for example) $3 \cdot 2x = x$, it is incorrect to then divide both sides by $x$ and end up with $6 = 1$ (you lose the solution $x = 0$ when you do this).
The correct next step is to subtract $\dfrac{2\tan x}{1-\tan^2 x}$ from both sides and continue as follows:
\begin{align}
  -\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} &= \frac{2\tan{x}}{1-\tan^2{x}}\\[0.3cm]
  -\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} - \frac{2\tan{x}}{1-\tan^2{x}}  &= 0\\[0.3cm]
  \frac{2\tan{x}}{1-\tan^2{x}} \left(-\tan^2 x - 1\right) &= 0
\end{align}
So then we have the following two equations:
$$\frac{2\tan x}{1-\tan^2 x} = 0 \qquad \text{or} \qquad -\tan^2 x - 1 = 0$$
The first equation is trivial.  The second one is perhaps even more so because the second one has no solutions.
