(1) I assume "log" on its own refers to the base-10 logarithm.
[tex]\left(e^{-2x}\log(\ln x)^3\right)'=\left(e^{-2x}\right)'\log(\ln x)^3+e^{-2x}\left(\log(\ln x)^3\right)'[/tex]
[tex]=-2e^{-2x}\log(\ln x)^3+\dfrac{e^{-2x}}{\ln10(\ln x)^3}\left((\ln x)^3\right)'[/tex]
[tex]=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10(\ln x)^3}\left(\ln x\right)'[/tex]
[tex]=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10\,x(\ln x)^3}[/tex]
[tex]=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}}{\ln10\,x\ln x}[/tex]
Note that writing [tex]\log(\ln x)^3=3\log(\ln x)[/tex] is one way to avoid using the power rule.
(2)
[tex]\left(e^{-2x}(\log(\ln x))^3\right)'=(e^{-2x})'(\log(\ln x))^3+e^{-2x}\left(\log(\ln x))^3\right)'[/tex]
[tex]=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2(\log(\ln x))'[/tex]
[tex]=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2\dfrac{(\ln x)'}{\ln10\,\ln x}[/tex]
[tex]=-2e^{-2x}(\log(\ln x))^3+\dfrac{3e^{-2x}(\log(\ln x))^2}{\ln10\,x\ln x}[/tex]
(3)
[tex]\left(\sin(xe^x)^3\right)'=\left(\sin(x^3e^{3x})\right)'=\cos(x^3e^{3x}(x^3e^{3x})'[/tex]
[tex]=\cos(x^3e^{3x})((x^3)'e^{3x}+x^3(e^{3x})')[/tex]
[tex]=\cos(x^3e^{3x})(3x^2e^{3x}+3x^3e^{3x})[/tex]
[tex]=3x^2e^{3x}(1+x)\cos(x^3e^{3x})[/tex]
(4)
[tex]\left(\sin^3(xe^x)\right)'=3\sin^2(xe^x)\left(\sin(xe^x)\right)'[/tex]
[tex]=3\sin^2(xe^x)\cos(xe^x)(xe^x)'[/tex]
[tex]=3\sin^2(xe^x)\cos(xe^x)(x'e^x+x(e^x)')[/tex]
[tex]=3\sin^2(xe^x)\cos(xe^x)(e^x+xe^x)[/tex]
[tex]=3e^x(1+x)\sin^2(xe^x)\cos(xe^x)[/tex]
(5) Use implicit differentiation here.
[tex](\ln(xy))'=(e^{2y})'[/tex]
[tex]\dfrac{(xy)'}{xy}=2e^{2y}y'[/tex]
[tex]\dfrac{x'y+xy'}{xy}=2e^{2y}y'[/tex]
[tex]y+xy'=2xye^{2y}y'[/tex]
[tex]y=(2xye^{2y}-x)y'[/tex]
[tex]y'=\dfrac y{2xye^{2y}-x}[/tex]
