в) $$ \frac{c}{3d} - \frac{4cd+c^2}{3d^2+3cd} = \frac{c}{3d} - \frac{c(4d+c)}{3d(d+c)} = \frac{c(d+c) - (4cd+c^2)}{3d(d+c)} = \frac{cd + c^2 - 4cd - c^2}{3d(d+c)} = \frac{-3cd}{3d(d+c)} = \frac{-c}{d+c} $$
Ответ: $$ \frac{-c}{d+c} $$