Answer : DCFL
Case 1 : If the Question has No Typo and is to be answered for what it is given.
i.e. $L_1 = \left \{ a^nb^k|n \neq k; n,k \geq 0 \right \}$ Which is DCFL.
and $L_2 = \left \{ xyx^R| x,y \in \left \{ 0,1 \right \}^+, \,\,\,and\,\,\, w^R \,\,\,stands\,\,\, for \,\,\,reversal\,\,\, of\,\,\, string\,\,\, w. \right \}$ Which is Regular and the Regular expression is $0(0+1)^+0 + 1(0+1)^+1.$
$L = \left \{ x^R| x \in L_1\,\,\, but\,\,\, not\,\,\, in\,\,\, L_2 ; and\,\,\, w^R \,\,\,stands\,\,\, for \,\,\,reversal\,\,\, of\,\,\, string\,\,\, w.\right \}$
Since $L_1, L_2 $ are defined over different Alphabets, there is No common string between them and so Set $L$ is nothing but Reversal of language $L_1.$ So,
$L = L_1^R = \left \{ b^ka^n|n \neq k; n,k \geq 0 \right \}$ , Which is DCFL as we know it.
Case 2 : If the Question has Typo and $L_2$ is defined over the same alphabet as $L_1$ i.e.
$L_1 = \left \{ a^nb^k|n \neq k; n,k \geq 0 \right \}$ Which is DCFL.
and $L_2 = \left \{ xyx^R| x,y \in \left \{ a,b \right \}^+, \,\,\,and\,\,\, w^R \,\,\,stands\,\,\, for \,\,\,reversal\,\,\, of\,\,\, string\,\,\, w. \right \}$ Which is Regular and the Regular expression is $a(a+b)^+a + b(a+b)^+b.$
$L = \left \{ x^R| x \in L_1\,\,\, but\,\,\, not\,\,\, in\,\,\, L_2 ; and\,\,\, w^R \,\,\,stands\,\,\, for \,\,\,reversal\,\,\, of\,\,\, string\,\,\, w.\right \}$
Now, If we look at $L_1,$ we can partition $L_1$ into the following languages : $a^+ , b^+, \left \{ a^nb^k|n \neq k; n,k \geq 1 \right \}$
Now, the Strings that are in $L_1$ But Not in $L_2$ are : $a,aa,b,bb, \left \{ a^nb^k|n \neq k; n,k \geq 1 \right \}$
So, $L$ is nothing but the Set of reversal of these strings i.e.
$L = \left \{ a,aa,b,bb \right \} \cup \left \{ b^ka^n|n \neq k; n,k \geq 1 \right \}$.. Which is DCFL as we know it. ($\left \{ b^ka^n|n \neq k; n,k \geq 1 \right \}$ is DCFL and we are adding finite number of strings to it which won't affect its language type.)
