$L=\mathop {\lim }\limits_{x \to 0}\frac{1-\sqrt[3]{cosx}}{sin^2x}-\frac{1-\sqrt{cosx}}{sin^2x}=\mathop {\lim }\limits_{x \to 0}(\frac{1-cosx}{sin^2x(1+\sqrt[3]{cosx}+\sqrt[3]{cos^2x})}-\frac{1-cosx}{sin^2x(1+\sqrt{cosx})})$$=\mathop {\lim }\limits_{x \to 0}(\frac{2sin^2\frac{x}{2}}{sin^2x(1+\sqrt[3]{cosx}+\sqrt[3]{cos^2x})}-\frac{2sin^2\frac{x}{2}}{sin^2x(1+\sqrt{cosx})})$
$=\mathop {\lim }\limits_{x \to 0}\frac{\frac{x^2}{2}.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^2.(\frac{sinx}{x})^2}(\frac{1}{1+\sqrt[3]{cosx}+\sqrt[3]{cos^2x}}-\frac{1}{1+\sqrt{cosx}})$
$=\mathop {\lim }\limits_{x \to 0}\frac{1}{2}(\frac{1}{3}-\frac{1}{2})=-\frac{1}{12}$