Let $ a = e^{ix}, $ $ b = e^{iy}, $ and $ c = e^{iz} $. Then \begin{align*} a + b + c &= e^{ix} + e^{iy} + e^{iz} \\ &= \cos x + i \sin x + \cos y + i \sin y + \cos z + i \sin z \\ &= (\cos x + \cos y + \cos z) + i (\sin x + \sin y + \sin z) \\ &= 0 \end{align*}Similarly, \begin{align*} \frac{1}{a} + \frac{1}{b} + \frac{1}{c} &= e^{-ix} + e^{-iy} + e^{-iz} \\ &= \cos x - i \sin x + \cos y - i \sin y + \cos z - i \sin z \\ &= (\cos x + \cos y + \cos z) - i (\sin x + \sin y + \sin z) \\ &= 0 \end{align*}Since $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0, $ $ \frac{ab + ac + bc}{abc} = 0, $ so \[ab + ac + bc = 0.\]Since $ a + b + c = 0, $ $ (a + b + c)^2 = 0, $ which expands as $ a^2 + b^2 + c^2 + 2(ab + ac + bc) = 0 $. Hence, \[a^2 + b^2 + c^2 = 0.\]But \begin{align*} a^2 + b^2 + c^2 &= e^{2ix} + e^{2iy} + e^{2iz} \\ &= \cos 2x + i \sin 2x + \cos 2y + i \sin 2y + \cos 2z + i \sin 2z, \end{align*}so $ \cos 2x + \cos 2y + \cos 2z = 0 $. Then \begin{align*} \cos 2x + \cos 2y + \cos 2z &= \cos^2 x - \sin^2 x + \cos^2 y - \sin^2 y + \cos^2 z - \sin^2 z \\ &= \frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x} + \frac{\cos^2 y - \sin^2 y}{\cos^2 y + \sin^2 y} + \frac{\cos^2 z - \sin^2 z}{\cos^2 z + \sin^2 z} \\ &= \frac{1 - \tan^2 x}{1 + \tan^2 x} + \frac{1 - \tan^2 y}{1 + \tan^2 y} + \frac{1 - \tan^2 z}{1 + \tan^2 z} \\ &= 0 \end{align*}It follows that \begin{align*} &(1 - \tan^2 x)(1 + \tan^2 y)(1 + \tan^2 z) \\ &\quad + (1 + \tan^2 x)(1 - \tan^2 y)(1 + \tan^2 z) \\ &\quad + (1 + \tan^2 x)(1 + \tan^2 y)(1 - \tan^2 z) = 0 \end{align*}Expanding, we get \begin{align*} &3 + \tan^2 x + \tan^2 y + \tan^2 z - (\tan^2 x \tan^2 y + \tan^2 x \tan^2 y + \tan^2 y \tan^2 z) \\ &\quad - 3 \tan^2 x \tan^2 y \tan^2 z = 0 \end{align*}Therefore, \begin{align*} &\tan^2 x + \tan^2 y + \tan^2 z - (\tan^2 x \tan^2 y + \tan^2 x \tan^2 z + \tan^2 y \tan^2 z) \\ &\quad - 3 \tan^2 x \tan^2 y \tan^2 z = \boxed{-3} \end{align*}