Find a closed formula for the following recursive function How can I express a closed-form formula for the following equation? 
$$f(n)=f(n-1)+\frac{C}{f(n-1)} $$
Where $C>0$ and $f(0)=\sqrt{C}$.

This is the best I can do. It seems like
\begin{align}
\frac{f(n)-f(n-1)}{n-(n-1)} = \frac{C}{f(n-1)}.
\end{align}
Let us solve the differential equation
\begin{align}
f' = \frac{C}{f} \ \ \implies \ \ \frac{d}{dx}(f)^2 = C
\end{align}
which means $f^2 = Cx$ or $f(x) = \sqrt{C}\sqrt{x}$. Observe
\begin{align}
\sqrt{C}\sqrt{n}-\sqrt{C}\sqrt{n-1} = \frac{\sqrt{C}}{\sqrt{n}+\sqrt{n-1}}
\end{align}
then it follows
\begin{align}
\frac{\sqrt{C}}{2f(n)}=\frac{\sqrt{C}}{2\sqrt{n}}\leq f(n)-f(n-1) \leq \frac{\sqrt{C}}{2\sqrt{n-1}} = \frac{\sqrt{C}}{2f(n-1)}
\end{align}
