1. Problem: If $f$ and $g$ are real functions with $f(x)=\frac{x-2}{x^2-3x+2}$ and $g(x)=x-3$, find $(f/g)(3)$. 2. Formula and rule: The quotient of two functions is given by $(f/g)(x)=\frac{f(x)}{g(x)}$ and it is defined only when $g(x)\neq 0$. 3. Simplify $f(x)$ by factoring the denominator: $x^2-3x+2=(x-1)(x-2)$. 4. Cancel common factors where allowed: $f(x)=\frac{x-2}{(x-1)(x-2)}=\frac{1}{x-1}$ for $x\neq 2$, noting a removable hole at $x=2$. 5. Evaluate at $x=3$: $f(3)=\frac{3-2}{3^2-3\cdot 3+2}=\frac{1}{2}$, and using the simplified form $f(3)=\frac{1}{3-1}=\frac{1}{2}$. 6. Evaluate $g(3)=3-3=0$. 7. Compute the quotient: $(f/g)(3)=\frac{f(3)}{g(3)}=\frac{1/2}{0}$ which is undefined because division by zero is not allowed. 8. Conclusion: The correct choice is (d) undefined.