Answer
$$
x^{\frac{2}{3}}+y^{\frac{2}{3}}=2; \quad\quad (1,1)
$$
The equation of the tangent line at the point $(1,1)$ is:
$$
y =-x+2
$$
Work Step by Step
$$
x^{\frac{2}{3}}+y^{\frac{2}{3}}=2; \quad\quad (1,1)
$$
Now, we can calculate $\frac{dy}{dx}$ by implicit differentiation,
$$
\begin{aligned}
\frac{2}{3} x^{-1 / 3}+\frac{2}{3} y^{-1 / 3} \frac{d y}{d x} &=0 \\ \frac{2}{3} y^{-1 / 3} \frac{d y}{d x} &=-\frac{2}{3} x^{-1 / 3} \\ \frac{d y}{d x} &=\frac{-\frac{2}{3} x^{-1 / 3}}{\frac{2}{3} y^{-1 / 3}} \\
&=-\frac{y^{1 / 3}}{x^{1 / 3}}
\end{aligned}
$$
To find the slope of the tangent line at the point $(1,1)$,
let $x=1$ and $y=1.$ The slope is
$$
\begin{aligned}
\frac{d y}{d x} &=-\frac{(1)^{1 / 3}}{(1)^{1 / 3}} \\
&=-1
\end{aligned}
$$
The equation of the tangent line is then found by using the point-slope form of the equation of a line.
$$
\begin{aligned}
(y-y_{1}) &=m(x-x_{1})\\
(y-(1)) &=(-1 )(x-(1))\\
y-1 &=-1 (x-1)\\
y &=-x+1+1\\
y &=-x+2\\
\end{aligned}
$$
Therefore, the equation of the tangent line at the point $(1,1)$ is:
$$
y =-x+2
$$
