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