The ring $L$ is a field (i.e. not a skew field) since $K$ is a field and you're only adjoining one (algebraic) element. Write $L=K[T]/(f(T))$ where $f$ is the minimal polynomial of $x$ over $K$. Note that $f$ has degree $\geq 2$ since $x\notin K$. Then $L\otimes_KL=L[T]/(f(T))$, and since $f(T)$ can be factored as $(T-x)^kg(T)$ in $L[T]$ with $x$ not a root of $g$, i.e., $T-x$ and $g(T)$ relatively prime in $L[x]$, the ring $L[T]/(f(T))$ is isomorphic to $L[T]/(T-x)^k\times L[T]/(g(T))$, which is not a domain, because either $k\geq 2$, in which case you have nilpotents, or $k=1$ and $g(T)\neq 1$, in which case you at least have zero divisors.
In general, if $L/K$ is non-trivial, then $L\otimes_KL$ is not a field. If $L$ is algebraic, pass to a finite subextension and use the argument above. If $L$ is not algebraic, then a theorem of Grothendieck says that the dimension of $L\otimes_KL$ has Krull dimension equal to the transcendence degree of $L$, so not zero (there is almost certainly a much easier way to see this, i.e., reducing to the case $K(T)\otimes_KK(T)$, but I'm not seeing it at the moment).
EDIT: Here's an argument that works in general for the assertion that $L\otimes_KL$ is not a field when $L/K$ is non-trivial. The argument was suggested in the comments by YACP. If $L\neq K$, then there exist distinct elements $\alpha\neq\beta\in L$ which are $K$-linearly independent. We may assume they are contained in a $K$-basis for $L$, in which case $1\otimes\alpha$ and $1\otimes\beta$ are $L$-linearly independent elements of $L\otimes_KL$. In particular, $\beta\otimes\alpha-\alpha\otimes\beta\neq 0$ in $L\otimes_KL$. Now, the $L$-linear multiplication map $x\otimes y\mapsto xy:L\otimes_KL\rightarrow L$ is non-zero ring map (since the target is non-zero). Since $\beta\otimes\alpha-\alpha\otimes\beta$ clearly lies in the kernel, we see that $L\otimes_KL$ has a non-trivial, proper ideal (namely the kernel). Therefore it is not a field.
It follows that $D\otimes_KL$ is not a skew field. If it were, then $L\otimes_KL$, being a commutative subring, would be a domain. But it isn't.