[tex]\bf 3-t\cfrac{dy}{dt}=8\implies 3-8=t\cfrac{dy}{dt}\implies \boxed{\cfrac{-5}{t}=\cfrac{dy}{dt}}\\\\
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n^2-\cfrac{ds}{dn}(6s-5)^{\frac{1}{3}}=0\implies n^2=\cfrac{ds}{dn}(6s-5)^{\frac{1}{3}}\implies \boxed{\cfrac{n^2}{(6s-5)^{\frac{1}{3}}}=\cfrac{ds}{dn}}[/tex]
nothing to it, is just a matter of solving for the derivative, which is already there
