Решение:
$$\frac{7x}{8x-3} - 3x = \frac{7x - 3x(8x-3)}{8x-3} = \frac{7x - 24x^2 + 9x}{8x-3} = \frac{16x - 24x^2}{8x-3}$$
Подставим $$x = \frac{1}{3}$$:
$$\frac{16(\frac{1}{3}) - 24(\frac{1}{3})^2}{8(\frac{1}{3})-3} = \frac{\frac{16}{3} - 24(\frac{1}{9})}{\frac{8}{3} - 3} = \frac{\frac{16}{3} - \frac{8}{3}}{\frac{8}{3} - \frac{9}{3}} = \frac{\frac{8}{3}}{-\frac{1}{3}} = -8$$
Ответ: -8