The correct option is:
$$(3,\sqrt{10})\cup [5,\sqrt{26})\cup \left\{6\right\}$$

Step-by-Step Explanation:

Let us analyze the given equation:
$$\left[x^2\right]={\left[x\right]}^2$$
for the domain defined by
$$3\le x\le x\le 6$$
where
$\left[x\right]$ represents the greatest integer function (or floor function) of x.

Let
$\left[x\right]=n$
where n is an integer. By definition of the greatest integer function:
$$n\le x<n+1$$
The given equation becomes:
$$\left[x^2\right]=n^2$$
Applying the definition of the greatest integer function again, this is equivalent to:
$$n^2\le x^2<n^2+1$$
Since
$x\ge 3>0$ , we can take the positive square root of all terms:
$$n\le x<\sqrt{n^2}$$

Now, we analyze each possible integer value of n in the domain
$3\le x\le 6$ :

1. For n = 3:
This corresponds to the interval
$3\le x<4$ .
Substituting n = 3 into our inequality:
$$3\le x<\sqrt{3^2}+1$$
which simplifies to:
$$3\le x<\sqrt{10}$$
Thus, the solution set in this interval is:
$$[3,\sqrt{10})$$

2. For n = 4:
This corresponds to the interval
$4\le x<5$ .
Substituting n = 4 into our inequality:
$$4\le x<\sqrt{4^2}+1$$
which simplifies to:
$$4\le x<\sqrt{17}$$
Thus, the solution set in this interval is:
$$[4,\sqrt{17})$$

3. For n = 5:
This corresponds to the interval
$5\le x<6$ .
Substituting n = 5 into our inequality:
$$5\le x<\sqrt{5^2}+1$$
which simplifies to:
$$5\le x<\sqrt{26}$$
Thus, the solution set in this interval is:
$$[5,\sqrt{26})$$

4. For x = 6:
At the boundary point x = 6:
$$[x{]}^2=6^2=36$$
and
$$\left[x^2\right]=\left[6^2\right]=36$$
Since both sides are equal,
$x=6$ is a valid solution, which gives:
$$\left\{6\right\}$$

Finding the Complete Solution Set (S):
Combining the solutions from all intervals, the total set S of all feasible values of x is:
$$S=[3,\sqrt{10})\cup [4,\sqrt{17})\cup [5,\sqrt{26})\cup \left\{6\right\}$$

Validating the Correct Option:
We need to identify which option is a subset of S.
Let's check the correctness of the option:
$$(3,\sqrt{10})\cup [5,\sqrt{26})\cup \left\{6\right\}$$
Since:
1. $(3,\sqrt{10})\subset [3,\sqrt{10})$
2. $[5,\sqrt{26})\subset [5,\sqrt{26})$
3. $\left\{6\right\}\subset \left\{6\right\}$
All parts of this union are subsets of S, making the entire set a valid subset of S.