Please help, functions!

To ensure that \( f \) is its own inverse, we need \( f(f(x)) = x \) for all \( x \). Given the piecewise definition of \( f(x) \):

\[
f(x) =
\begin{cases}
k(x) & \text{if } x > 3 \\
x^2 - 6x + 12 & \text{if } x \leq 3
\end{cases}
\]

Let's first analyze the case when \( x \leq 3 \).

### Case 1: \( x \leq 3 \)

\[
f(x) = x^2 - 6x + 12
\]

We need to calculate \( f(f(x)) \):

1. Compute \( f(x) \):

\[
f(x) = x^2 - 6x + 12
\]

2. Find \( f(f(x)) \):

We substitute \( f(x) \) back into the function:

\[
f(f(x)) = f(x^2 - 6x + 12)
\]

Now we need to check whether \( x^2 - 6x + 12 \) is greater than 3 for \( x \leq 3 \). We can evaluate this quadratic:

First, calculate the value when \( x = 3 \):

\[
f(3) = 3^2 - 6 \cdot 3 + 12 = 9 - 18 + 12 = 3
\]

For \( x < 3 \), we check the maximum value of \( x^2 - 6x + 12 \):

The expression \( x^2 - 6x + 12 \) is always minimized at \( x = 3 \), and since it is a quadratic opening upwards, \( x^2 - 6x + 12 \) will always be greater than or equal to \( 3 \).

Thus, for \( x < 3 \):
- \( f(x) = x^2 - 6x + 12 \) produces a value \( y = x^2 - 6x + 12 > 3 \)

Thus \( f(f(x)) \) will use the case \( k(y) \).

3. Given \( y = x^2 - 6x + 12 \):

We have:

\[
k(y) = k(x^2 - 6x + 12)
\]

And we want this to equal \( x \):

\[
k(x^2 - 6x + 12) = x
\]

### Case 2: Define \( k(y) \)

We already know that for \( k(y) \) where \( y = x^2 - 6x + 12 \):

We need \( k(y) \) when \( y > 3 \).

To find a suitable \( k(y) \), consider a simple equation with a variable transformation. Suppose we let \( k(y) = g(y) \), where we define \( g(y) \) such that \( f(g(y)) = y \).

Using the quadratic equation:

If \( y > 3 \):

Let \( k(y) \) be derived from \( k(x) \):

We need the outcome to solve \( g(y) = \sqrt{y} - 3\):

### Resulting function

Assuming \( y = k(x) \):

Then,

\[
k(y) = -3 + \sqrt{y - 3}
\]

### Summary of \( k(x) \)

Thus, we can derive:

\[
k(x) = -3 + \sqrt{x - 3} \quad \text{for } x > 3
\]

Thus, the function \( f(x) \) such that \( f \) is its own inverse is:

\[
f(x) =
\begin{cases}
-3 + \sqrt{x - 3} & \text{if } x > 3 \\
x^2 - 6x + 12 & \text{if } x \leq 3
\end{cases}
\]

Confirming:

- \( k(3) \) returns back confirming f is its own inverse,
- Further validations can ensure continuality and completeness over all functional parts.

Therefore, the final expression for \( k(x) \) is:

\[
\boxed{-3 + \sqrt{x - 3}} \quad \text{for } x > 3
\]