Answer
$y=-\dfrac{x}{2}+5.$
Work Step by Step
$\dfrac{d}{dx}(x^{\frac{2}{3}})+\dfrac{d}{dx}(y^{\frac{2}{3}})=\dfrac{d}{dx}(5)\rightarrow$
$\dfrac{2}{3\sqrt[3]{x}}$+$\dfrac{dy}{dx}(\dfrac{2}{3\sqrt[3]{y}})=0\rightarrow$
$\dfrac{dy}{dx}=-\dfrac{\sqrt[3]{y}}{\sqrt[3]{x}}$
At $(8, 1)\rightarrow\dfrac{dy}{dx}=-\dfrac{\sqrt[3]{1}}{\sqrt[3]{8}}=-\dfrac{1}{2}.$
Equation of tangent:
$(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y-1)=-\dfrac{1}{2}(x-8)\rightarrow y=-\dfrac{x}{2}+5.$
