Answer
$\text {The horizonal line that divides the area in equal two parts is}$
\begin{align}
y = \frac{9}{\sqrt[3] 4}
\end{align}
Work Step by Step
$\text {First, we have to find the area between y =x$^2$ and y = 9 as follows}$
$\text {The intersection between two functions:}$
\begin{align}
x^2 = 9 \\
x = \pm 3
\end{align}
$\text {Thus, the area is}$
\begin{align}
A = \int_{-3}^{3} 9 \ dx - \int_{-3}^{3} x^2 \ dx =54 -\left[\frac{x^3}{3} \right]_{-3}^{3} = 54-18 = 36
\end{align}
$\text {The half of this area is 18. Now we will use it to find y = k as follows}$
$\text {The intersection between two functions:}$
\begin{align}
x^2 = k \\
x = \pm \sqrt k
\end{align}
$\text {Thus, the area is}$
\begin{align}
A = 18 = \int_{-\sqrt k}^{\sqrt k} k \ dx -& \int_{-\sqrt k}^{\sqrt k} x^2 \ dx = 2k^{\frac{3}{2}} - \left[\frac{x^3}{3} \right]_{-\sqrt k}^{\sqrt k} = 2k^{\frac{3}{2}} - \frac{2k^{\frac{3}{2}}}{3} = \\
& = \frac{4k^{\frac{3}{2}}}{3} = 18 \Rrightarrow k = \frac{9}{\sqrt[3] 4}
\end{align}
$\text {Finally, the horizonal line that divides the area in equal two parts is}$
\begin{align}
y = \frac{9}{\sqrt[3] 4}
\end{align}
