Notice that since the linear term of the quadratic is $ -6, $ the vertex of the parabola that is the left side of $ f $ is at $ x=3 $. Therefore it might help to complete the square. \[x^2-6x+12=(x^2-6x+9)+3=(x-3)^2+3.\]We want to have that $ f(f(x))=x $ for every $ x $. Since $ f(f(3))=3, $ we know $ f $ is its own inverse at $ x=3, $ so we can restrict our attention to $ x\neq 3 $. Since $ f $ applied to any number less than $ 3 $ returns a number greater than $ 3, $ and we can get all numbers greater than $ 3 $ this way, applying $ f $ to any number greater than $ 3 $ must give a number less than $ 3 $. Therefore $ k(x)<3 $ for any $ x>3 $. If $ x>3 $ and $ f $ is its own inverse then \[x=f(f(x))=f(k(x))=3+\left(k(x)-3\right)^2,\]where in the last step we used that $ k(x)<3 $. Subtracting $ 3 $ from both sides gives \[\left(k(x)-3\right)^2 = x-3.\]Since we must have $ k(x) < 3, $ we know that $ k(x) - 3 $ is the negative number whose square is $ x-3 $. Therefore, we have $ k(x) - 3 = -\sqrt{x-3} $. Solving this for $ k(x) $ gives \[k(x)=\boxed{-\sqrt{x-3}+3}.\]