$3 \tan ^6 10^{\circ}-27 \tan ^4 10^{\circ}+33 \tan ^2 10^{\circ}=$

Let $x = \tan 10^{\circ}$ and consider the triple angle formula:
$\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$

For $\theta = 10^{\circ}$, we have $3\theta = 30^{\circ}$ and $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$, so:
$\frac{1}{\sqrt{3}} = \frac{3x - x^3}{1 - 3x^2}$

Squaring both sides to eliminate the radical:
$(1 - 3x^2)^2 = 3(3x - x^3)^2$

Expanding both sides gives:
$1 - 6x^2 + 9x^4 = 27x^2 - 18x^4 + 3x^6$

Rearranging all terms to one side yields:
$3x^6 - 27x^4 + 33x^2 - 1 = 0$

Therefore, $3x^6 - 27x^4 + 33x^2 = 1$

Final answer: $\boxed{1}$