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$\left(\frac{m^2-3m+9}{9m^2-1}\cdot \frac{3m^2+m}{m^3+27}-\frac{m+3}{3m^2-m}\right):\frac{5}{m^2+3m}-\frac{15m+6}{5-15m}=\frac{3}{5} \\\\$

$1) \frac{m^2-3m+9}{9m^2-1}\cdot \frac{3m^2+m}{m^3+27} = \frac{(m^2-3m+9)\cdot m(3m+1)}{(3m-1)(3m+1)(m+3)(m^2-3m+9)}=\frac{m}{(3m-1)(m+3)}$

2) $\frac{m}{(3m-1)(m+3)} - \frac{m+3}{m(3m-1)} =\frac{m^2-(m+3)^2}{m(3m-1)(m+3)} =\frac{(m-m-3)(m+m+3)}{m(3m-1)(m+3)} = \frac{-3(2m+3)}{m(3m-1)(m+3)}$

3) $\frac{-3(2m+3)}{m(3m-1)(m+3)}:\frac{5}{m(m+3)} = \frac{-3(2m+3)\cdot m(m+3)}{5m(3m-1)(m+3)}=\frac{-3(2m+3)}{5(3m-1)}$

$\frac{-3(2m+3)}{5(3m-1)} -\frac{3(5m+2)}{5(1-3m)} = \frac{-3(2m+3)}{5(3m-1)} +\frac{3(5m+2)}{5(3m-1)} = \frac{-3(2m+3)+3(5m+2)}{5(3m-1)}= \\\\ = \frac{3(-2m-3+5m+2)}{5(3m-1)} =\frac{3(3m-1)}{5(3m-1)} =\frac{3}{5}$