Помогите решить 35 балов

Решите задачу:

$(\frac{a+6}{3a+9}- \frac{1}{a+3})* \frac{3}{a-3}- \frac{6}{a^2-9}=\\\\=( \frac{a+6}{3(a+3)}- \frac{1}{a+3})* \frac{3}{a-3} - \frac{6}{(a-3)(a+3)}=\\\\= \frac{a+6-3}{3(a+3)}* \frac{3}{a-3}- \frac{6}{(a-3)(a+3)}=\\\\= \frac{a+3}{a+3}* \frac{1}{a-3}- \frac{6}{(a-3)(a+3)}=\\\\ \frac{a+3-6}{(a+3)(a-3)}= \frac{a-3}{(a+3)(a-3)} = \frac{1}{a+3}\\\\ a=- \frac{1}{4}\\\\ \frac{1}{- \frac{1}{4}+3 }= \frac{1}{\frac{11}{4}}=\frac{4}{11}$