9. (14 points) Find all real solutions of the equation $\frac{36}{\sqrt{x}}+\frac{9}{\sqrt{y}}$ $=42-9 \sqrt{x}-\sqrt{y}$, where if $x$ is a negative real number, then $\sqrt{x}=$ $\sqrt{|x|} \mathrm{i}$, with $\mathrm{i}$ being the imaginary unit.
9. Solution: By rearranging and completing the square, we can obtain:
$$
\frac{9}{\sqrt{x}}(\sqrt{x}-2)^{2}=\frac{-1}{\sqrt{y}}(\sqrt{y}-3)^{2}
$$

If $x, y$ are both positive real numbers, then both sides of the equation equal 0, at which point $x=4, y=9$;

If $x, y$ are both negative real numbers, then except for 42, all other terms in the original equation are purely imaginary, and the two sides of the equation cannot be equal;

If $x, y$ are one positive and one negative, comparing the original equation, if $x$ is negative, then $y$ is positive, at which point it is required that $\frac{36}{\sqrt{x}}+9 \sqrt{x}=0$, and $42-\frac{9}{\sqrt{y}}-\sqrt{y}=0$, solving gives $(x, y)=(-4,873 \pm 504 \sqrt{3})$; if $x$ is positive, then $y$ is negative, at which point it is required that $\frac{36}{\sqrt{x}}+9 \sqrt{x}=42$, and $-\frac{9}{\sqrt{y}}-\sqrt{y}=0$, solving gives $(x, y)=\left(\frac{62 \pm 14 \sqrt{13}}{9},-9\right)$.

In summary, the solutions to the equation are $(x, y)=(4,9),(-4, 873 \pm 504 \sqrt{3}),\left(\frac{62 \pm 14 \sqrt{13}}{9},-9\right)$.