How to show that this identity holds? By comparing the first terms of the Taylor expansion at $0$, it seems that
for $|x|<4/27$ the following identity holds:
$$\ln\left(\sum_{n=1}^{\infty}
\binom{3n}{n}\frac{x^{n-1}}{2n+1}\right)=
\sum_{n=1}^{\infty}
\binom{3n}{n}\frac{x^n}{n}.$$
I tried by differentiating both sides, but I am completely lost in a terrible mess. I wonder if there is a better strategy to handle it. Any idea?
 A: Let's try to use your idea of looking at derivatives: For $x=0$ both sides result in $0$, so indeed, if the derivatives are the same then we have the equality.
Let $g(x)$ be the left-hand side and $f(x)$ the right-hand side. The derivative of $g(x)$ is
$$g'(x)=\frac{1}{\sum_{n=1}^\infty\binom{3n}{n}\frac{x^{n-1}}{2n+1}}\sum_{n=2}^\infty\binom{3n}{n}\frac{(n-1)}{2n+1}x^{n-2}$$
and the derivative of the right-hand side is
$$f'(x)=\sum_{n=1}^\infty\binom{3n}{n}x^{n-1}$$
Just to simplify things, let's rewrite all these sums starting at $0$:
$$g'(x)=\frac{1}{\sum_{n=0}^\infty\binom{3n+3}{n+1}\frac{x^n}{2n+3}}\sum_{n=0}^\infty\binom{3n+6}{n+2}\frac{n+1}{2n+5}x^n$$
$$f'(x)=\sum_{n=0}^\infty\binom{3n+3}{n+1}x^n$$
So $g'(x)=f'(x)$ is equivalent to
$$\sum_{n=0}^\infty\binom{3n+6}{n+2}\frac{n+1}{2n+5}x^n=\left(\sum_{n=0}^\infty\binom{3n+3}{n+1}\frac{x^n}{2n+3}\right)\left(\sum_{n=0}^\infty\binom{3n+3}{n+1}x^n\right)$$
Now let's use the fact that $(\sum_{n=0}^\infty a_n)(\sum_{n=0}^\infty b_n)=\sum_{n=0}^\infty\left(\sum_{j=0}^n a_jb_{n-j}\right)$, so
\begin{align*}
\left(\sum_{n=0}^\infty\binom{3n+3}{n+1}\frac{x^n}{2n+3}\right)\left(\sum_{n=0}^\infty\binom{3n+3}{n+1}x^n\right)\hspace{-80pt}&\\
&=\sum_{n=0}^\infty\left(\sum_{j=0}^n\binom{3j+3}{j+1}\frac{x^j}{2j+3}\binom{3n-3j+3}{n-j+1}x^{n-j}\right)\\
&=\sum_{n=0}^\infty\left(\sum_{j=0}^n\frac{1}{2j+3}\binom{3j+3}{j+1}\binom{3n-3j+3}{n-j+1}\right)x^n.
\end{align*}
Finally, the problem is equivalent to showing that for all $n$,
$$\binom{3n+6}{n+2}\frac{1}{2n+5}=\sum_{j=0}^n\frac{1}{2j+3}\binom{3j+3}{j+1}\binom{3n-3j+3}{n-j+1}$$
For $n=0$ both sides yield $3$, so perhaps the rest will follow by induction.

EDIT: For $n=1$, the left-hand above side yields 12 but the right-hand side yields 24, so apparently this is not true (if there is no mistake in my arguments).
