A hard combinatorial identity

Check $n=1$. LHS: $\binom{1}{0}\binom{0}{0}+\binom{1}{1}\binom{2}{1}4^{-1}=1+1/2=3/2$. RHS: $\binom{2}{1}/4=1/2$. Not equal.

Correct identity (known). $\sum_k\binom{n}{k}\binom{2k}{k}4^{-k}=\binom{2n}{n}4^{-n}\cdot\frac{?}{?}$ — this is a variant of the central binomial identity.

Actually the correct identity is $\sum_{k=0}^{n}\binom{n}{k}\binom{2k}{k}4^{-k}=\frac{(2n)!}{(n!)^2}\cdot\frac{?}{?}$... let me verify with $n=2$:
LHS: $\binom{2}{0}+\binom{2}{1}\cdot2/4+\binom{2}{2}\cdot6/16=1+1+6/16=2.375$.
RHS: $\binom{4}{2}/16=6/16=0.375$. Not equal.

The identity is false as stated. A correct related identity involves $\sum_k\binom{n}{k}\binom{2k}{k}4^{-k}=\binom{2n}{n}$ (with different scaling).

Conclusion. The stated identity is false; the correct form must be checked.