Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
$x^{2 / 3}+y^{2 / 3}=4, \quad(-3 \sqrt{3}, 1) \quad (\text{astroid})$

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x^{\frac{2}{3}}+y^{\frac{2}{3}}=4

Find a derivative on both sides of equation with respect to $x$ . .

\begin{align*} \frac{d}{dx}\left(x^{\frac{2}{3}}+y^{\frac{2}{3}}\right)&=\frac{d}{dx}\left(4\right) &&\text{Use The Sum Rule.}\\ \frac{d}{dx}\left(x^{\frac{2}{3}}\right)+\frac{d}{dx}\left(y^{\frac{2}{3}}\right)&=\frac{d}{dx}\left(4\right) &&\text{Recall that $\left(c\right)'=0, c\in R$}\\ \frac{d}{dx}\left(x^{\frac{2}{3}}\right)+\frac{d}{dx}\left(y^{\frac{2}{3}}\right)&=0 &&\text{Use The Power Rule.}\\ \frac{2}{3}x^{\frac{2}{3}-1}+\frac{d}{dx}\left(y^{\frac{2}{3}}\right)&=0\\ \frac{2}{3}x^{-\frac{1}{3}}+\frac{d}{dx}\left(y^{\frac{2}{3}}\right)&=0 &&\text{Use The Chain Rule.}\\ \frac{2}{3}x^{-\frac{1}{3}}+\frac{d}{dy}\left(y^{\frac{2}{3}}\right)\frac{dy}{dx}&=0\\ \frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\cdot \frac{dy}{dx}&=0 &&\text{Divide both sides by $\frac{2}{3}$}\\ x^{-\frac{1}{3}}+y^{-\frac{1}{3}}\cdot \frac{dy}{dx}&=0 &&\text{Express $\frac{dy}{dx}$.}\\ \frac{dy}{dx}&=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}} &&\text{Simplify.}\\ \frac{dy}{dx}&=-\left(\frac{x}{y}\right)^{-\frac{1}{3}} &&\text{Subsitute $x=-3\sqrt{3}$, $y=1$}\\ \frac{dy}{dx}&=-\left(\frac{-3\sqrt{3}}{1}\right)^{-\frac{1}{3}}\\ \frac{dy}{dx}&=-\frac{1}{\sqrt[3]{-3\sqrt{3}}} &&\text{Simplify.}\\ \frac{dy}{dx}&=\frac{1}{\sqrt{3}} \end{align*}

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