NICE JOB. Your substitution gets rid of all the surds. Assuming 0 < cos(x) < 1
[MATH]cos(x) = t^6 \implies \sqrt{cos(x)} = t^3,\ \sqrt[3]{cos(x)} = t^2,\ cos^2(x) = t^{12},[/MATH]
[MATH]sin^2(x) = 1 - cos^2(x) = 1 - t^{12} = (1 - t^3)(1 + t^3)(1 + t^6) = (1 - t)(1 + t + t^2)(1 + t^3)(1 + t^6) \text { and}[/MATH]
[MATH]t^3 - t^2 = t^2((t = 1) = -\ t^2(1 - t).[/MATH]
[MATH]\therefore y = \dfrac{\sqrt{cos(x)} - \sqrt[3]{cos(x)}}{sin^2(x)} = \dfrac{t^3 - t^2}{1 - t^{12}} =[/MATH]
[MATH]\dfrac{-\ t^2 \cancel {(1 - t)}}{\cancel {(1 - t)}(1 + t + t^2)(1 + t^3)(1 + t^6)} = -\ \dfrac{t^2}{(1 + t + t^2)(1 + t^3)(1 + t^6)}.[/MATH]
[MATH]\lim_{x \rightarrow 0} cos(x) = 1 \implies \lim_{x \rightarrow 0} t^6 = 1 \implies \lim_{x \rightarrow 0} t^n =1 \implies[/MATH]
[MATH]\lim_{x \rightarrow 0} y = -\ \dfrac{1^2}{(1 + 1^1 + 1^2)(1 + 1^3)(1 + 1^6)} = -\ \dfrac{1}{3(2)(2)} = -\ \dfrac{1}{12}.[/MATH]
If you ask a machine just to give you the answer, the reason for the answer will not be apparent.