Relation between speed of sound c and compressibility
With your result

\begin{equation}
\beta_S=-\rho \frac{\partial \frac{1}{\rho}}{\partial p}
\end{equation}

you're nearly there. All that is missing is a little mathematical trick. Simply make the substitution $f = \frac{1}{\rho}$ and use the chain rule to differentiate $f$ by $\rho$ and you see that:

\begin{equation}
  \frac{\partial f}{\partial \rho} = - \frac{1}{\rho^2} \quad \Rightarrow \quad \mathrm{d} f = - \frac{1}{\rho^2} \mathrm{d} \rho \quad \Rightarrow \quad \mathrm{d} \frac{1}{\rho} = - \frac{1}{\rho^2} \mathrm{d} \rho \ .
\end{equation}

Using this and the definition of $c^2$ you get to the desired result:

\begin{equation}
\beta_S = -\rho \underbrace{\frac{\partial \frac{1}{\rho}}{\partial p}}_{= - \frac{\frac{1}{\rho^2} \partial \rho}{\partial p}} = 
\underbrace{(-\rho) (- \frac{1}{\rho^2})}_{= \frac{1}{\rho} } \underbrace{\frac{\partial \rho}{\partial p}}_{= \frac{1}{c^2}} \quad \Rightarrow \quad c^2 = \frac{1}{\rho \beta_S} \ .
\end{equation}