Final concentration of stock solution upon dilution
You are correct.  You might find it more intuitive to think of it this way:

In the end you will have $\pu{1 mL}$ of solution. In that end solution you have $\pu{0.1 mL} \times \frac{\pu{1mg}}{\pu{mL}}$ of stock, or (cancelling the $\pu{mL}$s) $\pu{0.1mg}$ of the chemical in question.  Therefore, final concentration is $\frac{\pu{0.1mg}}{\pu{1mL}} = 0.1 \frac{\pu{mg}}{\pu{mL}}$.