[tex]h(x)=(5-2x^6)^3+\dfrac1{5-2x^6}[/tex]
Let [tex]g(x)=5-2x^6[/tex] and [tex]f(x)=x^3+\dfrac1x[/tex]. Then [tex]h(x)=f(g(x))[/tex].
Set [tex]u=5-2x^6[/tex]. By the chain rule,
[tex]\dfrac{\mathrm dh}{\mathrm dx}=\dfrac{\mathrm dh}{\mathrm du}\cdot\dfrac{\mathrm du}{\mathrm dx}[/tex]
Since [tex]h(u)=u^3+\dfrac1u[/tex] and [tex]u(x)=5-2x^6[/tex], we have
[tex]\dfrac{\mathrm dh}{\mathrm du}=3u^2-\dfrac1{u^2}[/tex]
[tex]\dfrac{\mathrm du}{\mathrm dx}=-12x^5[/tex]
Then
[tex]\dfrac{\mathrm dh}{\mathrm dx}=\left(3u^2-\dfrac1{u^2}\right)(-12x^5)=\boxed{-12x^5\left(3(5-2x^6)^2-\dfrac1{(5-2x^6)^2}\right)}[/tex]
which we could rewrite slightly as
[tex]\dfrac{\mathrm dh}{\mathrm dx}=-\dfrac{12x^5(3(5-2x^6)^4-1)}{(5-2x^6)^2}[/tex]
