Find the sum of the squares of the values $x$ that satisfy $\frac{1}{x} + \frac{2}{x+3}+\frac{3}{x+6} = 1$.
1. Start with the given equation:
   \[
   \frac{1}{x} + \frac{2}{x+3} + \frac{3}{x+6} = 1
   \]

2. To eliminate the denominators, multiply both sides by \(x(x+3)(x+6)\):
   \[
   x(x+3)(x+6) \left( \frac{1}{x} + \frac{2}{x+3} + \frac{3}{x+6} \right) = x(x+3)(x+6) \cdot 1
   \]

3. Distribute the terms:
   \[
   (x+3)(x+6) + 2x(x+6) + 3x(x+3) = x(x+3)(x+6)
   \]

4. Simplify each term:
   \[
   (x^2 + 9x + 18) + 2x^2 + 12x + 3x^2 + 9x = x^3 + 9x^2 + 18x
   \]

5. Combine like terms:
   \[
   x^2 + 9x + 18 + 2x^2 + 12x + 3x^2 + 9x = x^3 + 9x^2 + 18x
   \]
   \[
   6x^2 + 30x + 18 = x^3 + 9x^2 + 18x
   \]

6. Rearrange the equation to set it to zero:
   \[
   x^3 + 9x^2 + 18x - 6x^2 - 30x - 18 = 0
   \]
   \[
   x^3 + 3x^2 - 12x - 18 = 0
   \]

7. Let the roots of the polynomial \(x^3 + 3x^2 - 12x - 18 = 0\) be \(a, b, c\). By Vieta's formulas, we know:
   \[
   a + b + c = -\frac{\text{coefficient of } x^2}{\text{leading coefficient}} = -\frac{3}{1} = -3
   \]
   \[
   ab + bc + ca = \frac{\text{coefficient of } x}{\text{leading coefficient}} = -12
   \]
   \[
   abc = -\frac{\text{constant term}}{\text{leading coefficient}} = -(-18) = 18
   \]

8. We need to find the sum of the squares of the roots:
   \[
   a^2 + b^2 + c^2
   \]

9. Use the identity:
   \[
   a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)
   \]

10. Substitute the known values:
    \[
    a^2 + b^2 + c^2 = (-3)^2 - 2(-12)
    \]
    \[
    a^2 + b^2 + c^2 = 9 + 24
    \]
    \[
    a^2 + b^2 + c^2 = 33
    \]

The final answer is \(\boxed{33}\).