Answer
$y={\frac{1}{\sqrt[2]{3}}}x+4$
Work Step by Step
$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4\\ \frac{2}{3}x^{-1/3}+\frac{2}{3}y^{-1/3}y'=0\\ \frac{2}{3\sqrt[3] x}+\frac{2y'}{3\sqrt[3] y}=0\\ -\frac{2}{3\sqrt[3] x}=\frac{2y'}{3\sqrt[3] y}\\ y'=-\sqrt[3]{\frac{y}{x}}$
Plug in $(-3\sqrt 3,1)$ in $y'$ to find the gradient
$y'=-\sqrt[3]{\frac{1}{-3\sqrt3}}\\ y'={\frac{1}{\sqrt[3]{3\sqrt3}}}$
$y'=\frac{1}{\sqrt[2]{3}}$
Equation of the tangent line at $(-3\sqrt 3,1)$
$y-1={\frac{1}{\sqrt[2]{3}}}(x-(-3\sqrt3))\\ y={\frac{1}{\sqrt[2]{3}}}x+4$
