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?

Why do you have an error?
If you divide though (by $\frac {2\tan x}{1-\tan^2x} = \tan 2x$ in this case), you must check that whatever you are dividing by does not equal 0.  Or, make a note to review the case that it does.
$\tan^2 x = -1$ or $\frac {2\tan x}{1-\tan^2x} = 0$
An alternatives you could have taken:
$0 = (1 - \tan^2 x)\frac {2\tan x}{1-\tan^2x}\\
0 = \sec^2 x \tan 2x$
But you are working a little bit too hard.
$\tan x = \frac {\tan x + \tan 2x}{1-\tan x\tan 2x}$
Rather than expand $\tan 2x$ right now, lets simplify a bit first
$\tan x- \tan^2 x \tan 2x = \tan x + \tan 2x\\
0 = \tan 2x (1+\tan^2 x)\\
0 = \tan 2x(\sec^2 x)$
Now, let's be even a little bit stupider.
$\tan x = \tan (x + k\pi)\\
\tan (x + k\pi) = \tan 3x\\
\tan^{-1}(\tan (x + k\pi)) = \tan^{-1}(\tan 3x)\\
x + k\pi = 3x\\
2x = k\pi$
