Binomial Identity I recently noted that 
$$\sum_{k=0}^{n/2} \left(-\frac{1}{3}\right)^k\binom{n+k}{k}\binom{2n+1-k}{n+1+k}=3^n$$
Is this a known binomial identity? Any proof or reference?

I rewrote your identity in an equivalent form 
$$
\sum_{k=0}^{n/2} (-1)^k 3^{n-k} \binom{n+k}{n, k} \binom{2n-k+1}{n-2k, n+k+1} = 3^{2n} ,
$$
and attempted to construct a proof by induction on $n$.  I did not succeed, but discovered an interesting hurdle: to 
prove the equally curious identity
$$
\sum_{k=0}^{n/2} (-1)^k 3^{n-k} \binom{n+k}{n, k} \binom{2n-k-2}{n-2k, n+k-2} = 0 .
$$
Does this latter identity have a human proof?
Here's a sketch of how I obtained the latter identity by rewriting the original equation.
By Pascal's rule we have
$
 \binom{n+k}{n, k}  =  \binom{n+k-1}{n-1, k} +  \binom{n+k-1}{n, k-1};
$
similarly
$
 \binom{2n-k+1}{n-2k, n+k+1}  =  3  \binom{2n-k-1}{n-2k-1, n+k} +  \binom{2n-k-2}{n-2k-3, n+k+1}  +  \binom{2n-k -2}{n-2k, n+k-2} .
$
Multiply these and regroup: the original summation now equals
$$
\sum_{k=0}^{n/2} (-1)^k 3^{n-k} \binom{n+k-1}{n-1, k} \ 3 \binom{2n-k-1}{n-2k-1, n+k} 
$$
$$
+ \sum_{k=0}^{n/2} (-1)^k 3^{n-k} \binom{n+k-1}{n, k-1} \ 3 \binom{2n-k-1}{n-2k-1, n+k} +
\sum_{k=0}^{n/2} (-1)^k 3^{n-k} \binom{n+k}{n, k}  \binom{2n-k-2}{n-2k-3, n+k+1}
$$
$$
+ \sum_{k=0}^{n/2} (-1)^k 3^{n-k} \binom{n+k}{n, k}  \binom{2n-k-2}{n-2k, n+k-2} .
$$
Evaluate this first summation by the induction hypothesis and obtain $9 \cdot 3^{2(n-1)}$, which equals $3^{2n}$
as desired.  The second and third summations are identical to each other except for a change of sign; together their sum is 0.
So our proof by induction (of the original statement) would conclude if we establish that this fourth summation equals 0.
