Is {wxw^r} a regular language?
$W^R$ means the reverse string of $W$.
You are right about the DFA. If a DFA $A$ accepts the language $L$ then it rejects all the words that are not in $L$. But the word $\sigma = 0110100$ is in the language!. Consider $W=0$ and $x=11010$.
As G.Bach eloquently stated in his comment(a very interesting property of the language):


$$\{WxW^R : W,x\in \{0,1\}^+\} = \{WxW : W\in\{0,1\} , x\in \{0,1\}^+\}$$

¿Why?

Because there is little restriction to $x$. We can consider that $x$ is almost the whole word with the exception of the first and the last symbol(because in the original language $|W|&gt;0$). The reverse string of any word with unitary size is the same word. So words like:

$$\sigma_1 = 01110000$$
$$\sigma_2 = 11111101$$ 

Are in the language. All that matter is the first and the last symbol. This language is very interesting because it appears to be irregular at the first sight.