Cho hàm số $y = f(x) =  - \frac{{{\rm{cos}}x}}{{3{{\sin }^3}x}} + \frac{4}{3}\cot x$.

$\begin{array}{l} f'\left( x \right) =  - \frac{{ - \sin x.3{{\sin }^3}x - \cos x.9{{\sin }^2}x.\cos x}}{{{{\left( {3{{\sin }^3}x} \right)}^2}}} - \frac{4}{3}.\frac{1}{{{{\sin }^2}x}}\\  = \frac{{{{\sin }^2}x + 3{{\cos }^2}x}}{{3{{\sin }^4}x}} - \frac{4}{{3{{\sin }^2}x}}\\  = \frac{{\cos 2x}}{{{{\sin }^4}x}} = \frac{{\cos 2x}}{{{{\left( {\frac{{1 - \cos 2x}}{2}} \right)}^2}}} = \frac{{4\cos 2x}}{{{{\left( {1 - \cos 2x} \right)}^2}}} \end{array}$

Do đó: $f'\left( {\frac{\pi }{3}} \right) = \frac{{4\cos \frac{{2\pi }}{3}}}{{{{\left( {1 - \cos \frac{{2\pi }}{3}} \right)}^2}}} =  - \frac{8}{9}$