Given the $n$th derivative of $f$ at $0$ find $f$. I want to find:
$$\frac{1}{2} \sum_{n=0}^{\infty} \frac{(1/6)^n(2n+2)!}{n!(n+1)!}$$
Where the sum looks like a Taylor series of $f(x)$ at $0$:
$$\sum_{n=0}^{\infty} \frac{x^n(2n+2)!}{n!(n+1)!}$$
Where:
$$f^{(n)}(0)=\frac{(2n+2)!}{(n+1)!}$$
Is there a technique to retreat $f(x)$?
 A: This is not really a general method, and I doubt you will find one. Finding a Taylor series is easy as it just requires us to do derivatives. However inverting $f^{(k)}(0)$ to find a closed form for $f(x)$ is hard (just like integration is usually much harder that differentiation) as it requires us to sum the Taylor series. The reason I choose the (seemingly arbitrary) method of considering $f^2(x)$ below is to take advantage of knowing the sum for $c_m$. There are many different methods one can use (complex analysis, taking advantage of known series, etc.). If you want to learn more I would suggest you look at some of the questions/answers about summing similar series on this site to learn different techniques one can use.

Since ${2n+2\choose n+1} = \frac{1}{n+1}\frac{(2n+2)!}{n!(n+1)!}$ your series can be written (after shifting the summation variable $n\to n-1$) $$\frac{1}{2}\sum_{n=1}^\infty {2n\choose n}n\left(\frac{1}{6}\right)^{n-1}$$
Now if we define $f(x) = \sum_{n=0}^\infty {2n\choose n}x^n$ then your sum is $\frac{f'(x)}{2}-1$ evaluated at $x=\frac{1}{6}$. To evaluate $f(x)$ consider
$$f^2(x) = \sum_{n=0}^\infty\sum_{k=0}^\infty {2n\choose n}{2k\choose k} x^{n+k} = \sum_{m=0}^\infty c_m x^m$$
where $c_m =$ $\sum_{n+k=m} {2n\choose n}{2k\choose k} = 4^m$ so $f^2(x) = \frac{1}{1-4x}$ for $|x| < \frac{1}{4}$ and $f(x) = \frac{1}{\sqrt{1-4x}}$ giving $3\sqrt{3}$ for your original series.

Here is an alternative derivation using complex analysis: by the binomial theorem the term ${2n\choose n}x^{n}$ is the coefficient of $z^{n}$ of the analytic function $x^{n}(1+z)^{2n}$. By the residue theorem it therefore follows that
$${2n\choose n}x^{n} = \frac{1}{2\pi i}\oint_{|z|=1} \frac{x^{n}(1+z)^{2n}}{z^{n}}\frac{{\rm d}z}{z}$$
Taking $f(x) = \sum_{n=0}^\infty {2n\choose n} x^n$ this gives us
$$f(x) = \frac{1}{2\pi i}\oint_{|z|=1}\sum_{n=0}^\infty\frac{x^{n}(1+z)^{2n}}{z^{n+1}}{\rm d}z = \frac{1}{2\pi i}\oint_{|z|=1}\frac{{\rm d}z}{z - x (1 + z)^2}$$
where we have assumed $|x|<\frac{1}{4}$ to have uniform convergence of the series allowing us to intechange the summation and integration. The integrand above have poles at $z_\pm = \frac{1-2 x \pm \sqrt{1-4 x}}{2 x}$ and only $z_-$ is inside $|z|=1$ so by the residue theorem
$$f(x) = \text{Res}\left[\frac{1}{z - x (1 + z)^2};z=\frac{1-2 x-\sqrt{1-4 x}}{2 x}\right] = \frac{1}{\sqrt{1-4x}}$$
This method is more general and can be used on several other series.
