problem: Two non-zero real numbers, $a$ and $b$, satisfy $ab=a-b$. Find the smallest possible value of: $\frac{a}{b}+\frac{b}{a}-ab$.
solution: Find the common denominator and replace the $ab$ in the numerator with $a-b$ to get
  
  \begin{align*}
  \frac{a}{b}+\frac{b}{a}-ab &= \frac{a^2+b^2-(ab)^2}{ab}\\
  &= \frac{a^2+b^2-(a-b)^2}{ab}\\
  &= \frac{a^2+b^2-(a^2-2ab+b^2)}{ab}\\
  &= \frac{2ab}{ab}=2.
  \end{align*}Hence the minimum possible value is the only possible value, $\boxed{2}$.