Wrong sign on the final answer, where did I go wrong? Algebra $12x^2-12x = 0$
This was suppose to be 18, but it is -18.   
I think it is because of how I modified one of the factors.
$\sqrt{x^2+x^2}$
$\sqrt{2x^2}$
$\sqrt{2}\cdot \sqrt{x^2}$
$x\sqrt{2}$
If I let wolfram do this, it gives the same but it says "assuming X is positive". Isn't that redundant? Any negative number squared is positive? The problem could be with the other factor that I modified.
$\sqrt{(6-x)^2+(6-x)^2}$
$x\sqrt2 -6\sqrt2$  
To sum it up:
$x\sqrt2(x\sqrt2-6\sqrt2) = 2x^2-12x$
It should be $-2x^2+12x$, because that would indeed give positive 18 as it's answer. 
Update! With all the info! 

The task is to find x such that the area of the rectangle would be as big as possible. I used the Pythagorean theorem and wrote out the following equation.
$\sqrt{(6-x)^2+(6-x)^2} \cdot \sqrt{x^2+x^2} = A$ of the rectangle.
The left factor would be the long side of the rectangle and the right factor would be the short side. I need to write this out nicely and then find when the derivative is 0. For max area. 

As you point out, by the Pythagorean Theorem, the area of the rectangle can be written as
$$\sqrt{2x^2}\sqrt{2(6-x)^2}.$$
Note that from the geometry, we have $0 < x < 6$.
It follows that $\sqrt{2x^2}=\sqrt{2}\;x$ and $\sqrt{2(6-x)^2}=\sqrt{2}\;(6-x)$.
Thus, if  $A(x)$ is the area, then
$$A(x)=2x(6-x).$$
(Incidentally, we can find the area in a more basic way. The two triangles at the Northeast and Southwest have combined area $x^2$, and the other two triangles have combined area $(6-x)^2$, for a total of $x^2+(6-x)^2=2x^2-12x+36$.  We can now minimize this, or subtract from $6^2$ and conclude that $A(x)=12x-2x^2$.) 
Maximize, using any of the standard tools. We could use the calculus, but in part because of the precalculus tag, we  complete the square. 
We have
$$A(x)=-2(x^2-6x)=-2((x-3)^2+9=18-2(x-3)^2.$$
The maximum is reached when $(x-3)^2=0$, that is, when $x=3$. The value of the area at $x=3$ is $18$.
Comment on error: The mistake in the post was in the simplification of $\sqrt{(6-x)^2+(6-x)^2}$.
You obtained $x\sqrt{2} -6\sqrt{2}$, and the correct value is $6\sqrt{2}-x\sqrt{2}$. The expression $x\sqrt{2}-6\sqrt{2}$ cannot be right, since for $\;0<x <6$, it gives a negative number.  Maybe if you had written it as $\sqrt{2}(x-6)$ the mistake would have become obvious. 
We are trying to find a certain length, the square root of $2(6-x)^2$.  The square root of any (non-negative) number is non-negative. 
A rule that one might want to remember is that $\sqrt{a^2}=|a|$. So our square root is $\sqrt{2}|6-x|$. Since $0 \le x \le 6$ by the geometry, $|6-x|=6-x$.
