Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 28

Find an equation of the tangent to the curve $x^{\frac{2}{3}} + y^{\frac{2}{3}} = 4$ at the point $\displaystyle \left(-3\sqrt{3},1\right)$ using Implicit Differentiation.
If $y'= m\text{ (slope)}$ then,
$\displaystyle \frac{d}{dx} \left(x^{\frac{2}{3}}\right) + \frac{d}{dx} \left(y^{\frac{2}{3}}\right) = \frac{d}{dx} (4)$


$
\begin{equation}
\begin{aligned}
\frac{2}{3} (x)^{\frac{-1}{3}} + \frac{2}{3} (y)^{\frac{-1}{3}} \frac{dy}{dx} &= 0\\
\\
\frac{2}{3\sqrt[3]{x}} + \frac{2}{3 \sqrt[3]{y}} y' &= 0\\
\\
\frac{2}{3\sqrt[3]{y}} y' &= \frac{-2}{3 \sqrt[3]{x}}\\
\\
y' &= \frac{-2(3\sqrt[3]{y})}{2(3\sqrt[3]{x})}\\
\\
y' &= \frac{-\sqrt[3]{y}}{\sqrt[3]{x}}
\end{aligned}
\end{equation}
$

For $x = -3\sqrt{3}$ and $y =1 $, we obtain

$
\begin{equation}
\begin{aligned}
y' = m &= - \frac{\sqrt[3]{y}}{\sqrt[3]{x}}\\
\\
m &= - \frac{\sqrt[3]{1}}{\sqrt[3]{3-\sqrt{3}}}\\
\\
m &= \frac{1}{\sqrt{3}}
\end{aligned}
\end{equation}
$


Using the point slope form

$
\begin{equation}
\begin{aligned}
y - y_1 &= m(x-x_1)\\
\\
y - 1 & = \frac{1}{\sqrt{3}} [x - ( -3 \sqrt{3})]\\
\\
y - 1 & = \frac{1}{\sqrt{3}} (x + 3\sqrt{3})\\
\\
y &= \frac{x+3\sqrt{3}}{\sqrt{3}} +1 \\
\\
y &= \frac{x+3\sqrt{3}+\sqrt{3}}{\sqrt{3}}\\
\\
y &= \frac{x+4\sqrt{3}}{\sqrt{3}} && \text{Equation of the tangent line at } (-3\sqrt{3},1)
\end{aligned}
\end{equation}
$