1. **State the problem.** We need to simplify the expression $$\frac{x^2-25}{x^2+5x}\div\frac{xy+6x-5y-30}{5x-15}$$ 2. **Rewrite division as multiplication by the reciprocal.** For fractions, the rule is $$\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\cdot\frac{d}{c}$$ So we get $$\frac{x^2-25}{x^2+5x}\cdot\frac{5x-15}{xy+6x-5y-30}$$ 3. **Factor every polynomial.** Factor the first numerator: $$x^2-25=(x-5)(x+5)$$ Factor the first denominator: $$x^2+5x=x(x+5)$$ Factor the second numerator: $$5x-15=5(x-3)$$ Factor the second denominator by grouping: $$xy+6x-5y-30=x(y+6)-5(y+6)=(x-5)(y+6)$$ So the expression becomes $$\frac{(x-5)(x+5)}{x(x+5)}\cdot\frac{5(x-3)}{(x-5)(y+6)}$$ 4. **Cancel common factors carefully.** We can cancel $x+5$ and $x-5$ because they appear in both numerator and denominator. $$\frac{(x-5)\cancel{(x+5)}}{x\cancel{(x+5)}}\cdot\frac{5(x-3)}{\cancel{(x-5)}(y+6)}$$ After canceling, we have $$\frac{1}{x}\cdot\frac{5(x-3)}{y+6}$$ 5. **Multiply the remaining factors.** $$\frac{1}{x}\cdot\frac{5(x-3)}{y+6}=\frac{5(x-3)}{x(y+6)}$$ 6. **Final answer.** $$\boxed{\frac{5(x-3)}{x(y+6)}}$$