\lim_{x\rightarrow 0}\left(\frac 1{\sin^2x}-\frac 1{x^2}\right)=\lim_{x\rightarrow 0}\frac{x^2-\sin^2x}{x^2\sin^2x}=\lim_{x\rightarrow 0}\frac{x-\sin x}{x^3}\cdot\frac{x}{\sin x}\cdot\frac{x+\sin x}{\sin x}=2\lim_{x\rightarrow 0}\frac{x-\sin x}{x^3}