I think factorizing everything you can first will make the simplification ... well, simpler.
[tex]\dfrac{x^2 - 3x}{x^2 + 13x + 36} \div \dfrac{x^2+7x}{x^2+16x+63} = \dfrac{x(x-3)}{(x+4)(x+9)} \div \dfrac{x(x+7)}{(x+7)(x+9)}[/tex]
The factors of [tex]x+7[/tex] in the second rational expression cancel:
[tex]\dfrac{x^2 - 3x}{x^2 + 13x + 36} \div \dfrac{x^2+7x}{x^2+16x+63} = \dfrac{x(x-3)}{(x+4)(x+9)} \div \dfrac{x}{x+9}[/tex]
Now, use the property
[tex]\dfrac ab \div \dfrac cd = \dfrac ab \times \dfrac dc[/tex]
(this is the property of multiplication having to do with multiplicative inverse, or "inverting the divisor" as the question calls it) to write
[tex]\dfrac{x^2 - 3x}{x^2 + 13x + 36} \div \dfrac{x^2+7x}{x^2+16x+63} = \dfrac{x(x-3)}{(x+4)(x+9)} \times \dfrac{x+9}x[/tex]
and we see some more cancellation, namely of the factors of [tex]x[/tex] and [tex]x+9[/tex].
[tex]\dfrac{x^2 - 3x}{x^2 + 13x + 36} \div \dfrac{x^2+7x}{x^2+16x+63} = \boxed{\dfrac{x-3}{x+4}}[/tex]
