Find the particular solution of the differential equation $\sqrt{x} + \sqrt{y}y' =0$, with $y(1) = 9$.
I get:
$\sqrt{y}dy = -\sqrt{x}dx \Rightarrow \frac{2}{3}y^{\frac{3}{2}} = -\frac{2}{3}x^{\frac{3}{2}}+C$.
What should I do from here?

Solve it by separation of variables because it's a separable differential equation:
$$
\sqrt{x} + \sqrt{y}y' =0\\
\sqrt{y}\frac{dy}{dx}=-\sqrt{x}\\
\int\sqrt{y}\frac{dy}{dx}\,dx=-\int\sqrt{x}\,dx\\
\int\sqrt{y}\,dy=-\int\sqrt{x}\,dx\\
\frac{2\sqrt{y^3}}{3}+C_1=-\frac{2\sqrt{x^3}}{3}+C_2\\
\sqrt{y^3}=-\sqrt{x^3}+C\\
y=\sqrt[3]{(C-\sqrt{x^3})^2}
$$
Now that we have found the general solution, we can use the initial condition $y(1)=9$ to find a particular solution:
$$
y(1)=9\implies \sqrt[3]{(C-\sqrt{1^3})^2}=9\implies C=28
$$
Your final answer is $y=\sqrt[3]{(28-\sqrt{x^3})^2}$.
