Recall that we can write
[tex]f(x)^{g(x)}=e^{\ln f(x)^{g(x)}}=e^{g(x)\ln f(x)}[/tex]
so that when we take the derivative, we get by the chain rule
[tex]\left(f(x)^{g(x)}\right)'=(g(x)\ln f(x))' e^{g(x)\ln f(x)}=(g(x)\ln f(x))'f(x)^{g(x)}[/tex]
Here, we have [tex]f(x)=3-\sin2x[/tex] and [tex]g(x)=\sqrt[3]{x}=x^{1/3}[/tex]. By the product rule,
[tex]\left(x^{1/3}\ln(3-\sin2x)\right)'=\left(x^{1/3}\right)\ln(3-\sin2x)+x^{1/3}(\ln(3-\sin2x))'[/tex]
[tex]=\dfrac13x^{-2/3}\ln(3-\sin2x)+x^{1/3}\dfrac{(3-\sin2x)'}{3-\sin2x}[/tex]
[tex]=\dfrac13x^{-2/3}\ln(3-\sin2x)+x^{1/3}\dfrac{2\cos2x}{\sin2x-3}[/tex]
[tex]=\dfrac1{3x^{2/3}}\left(\ln(3-\sin2x)+3x\dfrac{2\cos2x}{\sin2x-3}\right)[/tex]
[tex]=\dfrac1{3\sqrt[3]{x^2}}\left(\ln(3-\sin2x)+\dfrac{6x\cos2x}{\sin2x-3}\right)[/tex]
So we have
[tex]y'=\dfrac{(3-\sin2x)^{\sqrt[3]{x}}}{3\sqrt[3]{x^2}}\left(\ln(3-\sin2x)+\dfrac{6x\cos2x}{\sin2x-3}\right)[/tex]
