[tex]g(x)=\dfrac{\cos x}{f(x)}[/tex]
[tex]\implies g'(x)=-\dfrac{f(x)\sin x+f'(x)\cos x}{f(x)^2}[/tex]
You know that [tex]f\left(\dfrac\pi3\right)=4[/tex] and [tex]f'\left(\dfrac\pi3\right)=-2[/tex], so
[tex]g'\left(\dfrac\pi3\right)=-\dfrac{4\sin\frac\pi3-2\cos\frac\pi3}{4^2}=\dfrac{1-2\sqrt3}{16}[/tex]
