$L$ = {$ab,aabb,aaabbb,aaaabbbb....... $} $\Rightarrow DCFL$
$R$ = {$\epsilon, a ,b , aa,bb,ab,ba,aaa,bbb ,aabb,aaaabbbb,,,,,,,,$} $\Rightarrow Regular$
$L\cup R$ = {$\epsilon, a ,b , aa,bb,ab,ba,aaa,bbb ,aabb,aaaabbbb,,,,,,,,$} = $R$ $\Rightarrow Regular$
The thing is $R$ is a set of all strings over {$a,b$}. So $union$ of $R$ and $L$ is $R$ only. So it's regular.
$DCFL \cup Regular$ = $DCFL$ and Regular languages are subset of DCFL .