\[\frac{1}{3x + 1} - \frac{1}{9x^{2} + 6x + 1} = 2\]
\[ОДЗ:\ \ x \neq - \frac{1}{3}\]
\[\frac{3x + 1 - 1}{9x^{2} + 6x + 1} = 2\]
\[3x = 2 \cdot \left( 9x^{2} + 6x + 1 \right)\]
\[3x = 18x^{2} + 12x + 2\]
\[18x^{2} + 12x + 2 - 3x = 0\]
\[18x^{2} + 9x + 2 = 0\]
\[D = b^{2} - 4ac =\]
\[= 81 - 4 \cdot 18 \cdot 2 =\]
\[= 81 - 144 < 0\ \]
\[Ответ:\ \ нет\ решений.\]