In the second fraction, factorize the denominator:
[tex]x^2-9x-10=(x-10)(x+1)[/tex]
Then we have, as long as [tex]x\neq10[/tex],
[tex]\dfrac1{x-10}\div\dfrac{9x}{x^2-9x-10}=\dfrac{\frac1{x-10}}{\frac{9x}{(x-10)(x+1)}}=\dfrac1{\frac{9x}{x+1}}=\dfrac{x+1}{9x}[/tex]
which you could also write as
[tex]\dfrac{x+1}{9x}=\dfrac x{9x}+\dfrac1{9x}=\dfrac19+\dfrac1{9x}[/tex]
