We need to evaluate $L=\lim_{x\to 0}\frac{\sin^{-1}x-\tan^{-1}x}{3x^3}.$

Use the standard Maclaurin expansions near $x=0$: $\sin^{-1}x = x+\frac{x^3}{6}+O(x^5),$ $\tan^{-1}x = x-\frac{x^3}{3}+O(x^5).$

Subtract the two series: $\sin^{-1}x-\tan^{-1}x = \left(x+\frac{x^3}{6}\right)-\left(x-\frac{x^3}{3}\right)+O(x^5).$ The $x$ terms cancel, so $\sin^{-1}x-\tan^{-1}x = \frac{x^3}{6}+\frac{x^3}{3}+O(x^5).$ $=\frac{x^3}{2}+O(x^5).$

Substitute into the limit: L=\lim_{x\to 0}\frac{\frac{x^3}{2}+O(x^5)}{3x^3}=rac{1/2}{3}=\frac{1}{6}.

Now compute: $6L+1=6\left(\frac{1}{6}\right)+1=1+1=2.$

Hence the correct option is $\boxed{\text{D: }2}.$

Comparison with stored answer: Stored correct answer is D, which matches our result.