We start with the indifference curve for $k=2$. We determine the curve that connects all the points $(x,y)$ such that
$$\begin{align}
2x^{\tfrac{1}{3}}y^{\tfrac{2}{3}} &= 2\\
x^{\tfrac{1}{3}}y^{\tfrac{2}{3}} &= 1\\\
y^{\tfrac{2}{3}} &= x^{-\tfrac{1}{3}}\\
y &= \left(x^{-\tfrac{1}{3}}\right)^{\tfrac{3}{2}} = x^{-\tfrac{1}{2}} = \dfrac{1}{\sqrt{x}}.
\end{align}$$
The second indifference curve, with value $k=8$, is determined in the same way. We determine the curve that connects all the points $(x,y)$ such that
$$\begin{align}
2x^{\tfrac{1}{3}}y^{\tfrac{2}{3}} &= 8\\
x^{\tfrac{1}{3}}y^{\tfrac{2}{3}} &= 4\\\
y^{\tfrac{2}{3}} &= 4x^{-\tfrac{1}{3}}\\
y &= \left(4x^{-\tfrac{1}{3}}\right)^{\tfrac{3}{2}}=4^{\tfrac{3}{2}}\left(x^{-\tfrac{1}{3}}\right)^{\tfrac{3}{2}} = 8x^{-\tfrac{1}{2}} = \dfrac{8}{\sqrt{x}}.
\end{align}$$
In the coordinate system below both indifference curves are shown.
We consider the function $U(x,y) =2x^{\tfrac{1}{3}}y^{\tfrac{2}{3}}$. The graph of this function can be found here: Cobb-Douglas functions. Here we draw the indifference curves of $U(x,y)$ with values $k=2$ and $k=8$.