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[tex]\displaystyle x=\frac{2\pi }{3}\\\\f(x)=\frac{2x-3}{sinx}\ \ ,\qquad \qquad \Big(\frac{u}{v}\Big)'=\frac{u'v-uv'}{v^2}\\\\\\f'(x)=\frac{(2x-3)'\cdot sinx-(2x-3)\cdot (sinx)'}{sin^2x}=\frac{2\, sinx-(2x-3)\, cosx}{sin^2x}\\\\\\f'\Big(\frac{2\pi }{3}\Big)=\frac{2\, sin\dfrac{2\pi}{3}-\Big(\dfrac{4\pi}{3}-3\Big)\cdot cos\dfrac{2\pi}{3}}{sin^2\dfrac{2\pi}{3}} =\frac{2\cdot \dfrac{\sqrt3}{2}+\dfrac{4\pi -9}{3}\cdot \dfrac{1}{2}}{\dfrac{3}{4}}=[/tex]

[tex]\displaystyle =\frac{6\sqrt3+4\pi -9}{6\cdot \dfrac{3}{4}}=\frac{2(4\pi +6\sqrt3-9)}{9}[/tex]