Which solution is correct for $\tan x+\tan2x+\tan3x=\tan x\tan2x\tan3x$? 
Equation to solve is:
$$\tan x+\tan2x+\tan3x=\tan x\tan2x\tan3x$$

$\text {Solution 1:}$
$$\tan x+\tan 2x=\tan x \tan 2x\tan 3x-\tan 3x$$
$$\tan x+\tan 2x=\tan 3x(\tan x \tan 2x-1)$$
$$\tan x+\tan 2x=-\tan 3x(1-\tan x \tan 2x)$$
$$\dfrac{\tan x+\tan 2x} {(1-\tan x\tan 2x)}= -\tan 3x$$
$$\tan 3x=-\tan 3x$$
$$2\tan 3x=0$$
$$\tan 3x=0$$
$$3x=n\pi, \text {where n is an integer}$$
$$x=\frac{n\pi}{3}$$
$\text {Solution 2:}$
$$\tan x+\tan 2x=\tan x\tan 2x \tan 3x-\tan 3x$$
$$\tan x+\tan 2x=\tan 3x(\tan x \tan 2x-1)$$
$$\tan x+\tan 2x=-\tan 3x(1-\tan x \tan 2x)$$
$$ \frac{\tan x+\tan 2x} {(1-\tan x \tan 2x)}= -\tan 3x$$
$$\tan 3x=-\tan 3x$$
$$\tan 3x=\tan(-3x)$$
$$3x=n\pi + (-3x)$$
$$6x=n\pi, \text  {where n is an integer}$$
$$x=\frac{n\pi}{6}$$
Also,$\dfrac{n\pi}{3} \subset \dfrac{n\pi}{6}, \text {i.e. Solution 1 is contained in Solution 2.}$
But, the extra solution provided $\text {Solution 2}$ (which are not present in $\text {Solution 1}$) are of no use, as the original equation is undefined at those point, for eg, $\frac{\pi}{6}$ and $\tan$ function is not defined at $\frac{3\pi}{6}$. So, $\text {Solution 1}$ can give all the solutions for the equation.
My book has $\text {Solution 2}$ as the answer. Which is the correct answer?

Whenever $n$ is odd, $\tan \dfrac {3n\pi}6$ is undefined.
Therefore you should choose solution 1 (it is more intuitive anyway)
On an unrelated note, add a backslash "\" before each $tan$.
