We have done this before in Example 4 at level curves; here we will apply the rule in tangent line to a level curve. We will see that this gives the same answer.
$$\begin{align}
\dfrac{U'_x(x,y)}{U'_y(x,y)}= \dfrac{\frac{2}{3}x^{-\frac{2}{3}}}{\frac{4}{3}y^{-\frac{2}{3}}} .
\end{align}$$
$$\begin{align}
\textrm{slope} &=- \dfrac{\frac{2}{3}\cdot(1)^{-\frac{2}{3}}}{\frac{4}{3}\cdot(1)^{-\frac{2}{3}}}\\
& = - \frac{1}{2}.
\end{align}$$
The tangent line is in general given by $t(x)=ax+b$. Hence, it holds that $a=-\frac{1}{2}$.
In order to determine $b$ we use that the tangent line goes through $(x,y)=(1,1)$. Hence, $t(1)=-\frac{1}{2}\cdot 1 + b=1$. This gives $b=\frac{3}{2}$.
Hence, the tangent line is given by $t(x)=-\frac{1}{2}x+\frac{3}{2}$.
Consider the function $$U(x,y) =2x^{\tfrac{1}{3}}y^{\tfrac{2}{3}}.$$ Determine the tangent line to the indifference curve of $U(x,y)$ at $(x,y)=(1,1)$.