When [tex]x=8[/tex], the function [tex]x^{2/3}[/tex] takes on a value of [tex]8^{2/3}=\sqrt[3]{64}=4[/tex]. So you're looking to find the tangent line to [tex]x^{2/3}[/tex] through the point [tex](8,4)[/tex].
The tangent line through this point has the value of the derivative of [tex]x^{2/3}[/tex] when [tex]x=8[/tex]. The derivative is, by the power rule,
[tex]\dfrac{\mathrm d}{\mathrm dx}x^{2/3}=\dfrac23 x^{-1/3}\stackrel{x=8}\implies\text{slope}=\dfrac23\times8^{-1/3}=\dfrac13[/tex]
The point-slope form of the tangent line is
[tex]y-4=\dfrac13(x-8)\implies L(x)=y=\dfrac13x+\dfrac43[/tex]
