Answer
$$y =-x+2$$
Work Step by Step
Given $$x^{2 / 3}+y^{2 / 3}=2,\ \ \ \ (1,1) $$
Differentiate both sides with respect to $x$
\begin{align*}
\frac{d}{dx}(x^{2 / 3}+y^{2 / 3})&=\frac{d}{dx}(2)\\
\frac{2}{3} x^{-\frac{1}{5}}+\frac{2}{3} y^{-\frac{1}{5}} y^{\prime}&=0\\
\frac{2}{3} y^{-\frac{1}{3}} y^{\prime}&=-\frac{2}{3} x^{-\frac{1}{3}}\\
y'&= \frac{-x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}
\end{align*}
Then at $(2,1)$
\begin{align*}
m&= \frac{-1^{-\frac{1}{3}}}{1^{-\frac{1}{3}}}\\
&=- 1
\end{align*}
Then the tangent line is given by
\begin{align*}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-1}{x-1}&=-1\\
y-1 &= -x+1\\
y&=-x+2
\end{align*}
