Use the chain rule:
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dx}[/tex]
We have
[tex]y=8u-6\implies\dfrac{\mathrm dy}{\mathrm du}=8[/tex]
[tex]u=3x-8\implies\dfrac{\mathrm du}{\mathrm dx}=3[/tex]
so we get
[tex]\dfrac{\mathrm dy}{\mathrm dx}=8\cdot3=\boxed{24}[/tex]
Alternatively, you can substitute u in the definition of y and differentiate with respect to x :
[tex]y=8u-6=8(3x-8)=24x-64\implies\dfrac{\mathrm dy}{\mathrm dx}=24[/tex]
