Prove that $\sum_{l=0}^{n} \binom{n}{l}^2 (x+y)^{2l} (x-y)^{2(n-l)} = \sum_{l=0}^{n} \binom{2l}{l} \binom{2(n-l)}{n-l} x^{2l}y^{2(n-l)}$

The RHS is the coefficient of $z^{2n}$ in the product between $$ \sum_{l\geq 0}\binom{2l}{l}(xz)^{2l}=\frac{1}{\sqrt{1-4x^2 z^2}}\qquad\text{and}\qquad \sum_{l\geq 0}\binom{2l}{l}(yz)^{2l}=\frac{1}{\sqrt{1-4y^2 z^2}} $$ i.e. the coefficient of $z^{2n}$ in $\frac{1}{\sqrt{(1-4x^2 z^2)(1-4y^2 z^2)}}$.
In the LHS we may recognize a Legendre polynomial: $$ P_n(t)=\frac{1}{2^n}\sum_{l=0}^{n}\binom{l}{n}^2(t+1)^l (t-1)^{n-l} $$ hence the given identity turns out to be a consequence of the following identity, about the generating function for Legendre polynomials: $$ \frac{1}{\sqrt{1-2xt+t^2}}=\sum_{n\geq 0}t^n P_n(x) $$ which can be proved through Rodrigues' formula or Bonnet's recursion formula.

We seek to verify that

$$\sum_{l=0}^{n} {n\choose l}^2 (x+y)^{2l} (x-y)^{2n-2l} = \sum_{l=0}^{n} {2l\choose l} {2n-2l\choose n-l} x^{2l}y^{2n-2l}.$$

Now we see on the LHS that the powers of $x$ and $y$ always add up to $2n$ and the exponent on $x$ determines the one on $y.$ Extracting the coefficient on $[x^q][y^{2n-q}]$ we obtain

$$\sum_{l=0}^{n} {n\choose l}^2 \sum_{p=0}^q {2l\choose p} (-1)^{2n-2l-(q-p)} {2n-2l\choose q-p} \\ = \sum_{l=0}^{n} {n\choose l}^2 \sum_{p=0}^q {2l\choose p} (-1)^{q-p} [z^{q-p}] (1+z)^{2n-2l} \\ = [z^q] (-1)^q \sum_{l=0}^{n} {n\choose l}^2 (1+z)^{2n-2l} \sum_{p=0}^q {2l\choose p} (-1)^p z^p.$$

We may extend $p$ to infinity because with $p\gt q$ there is no contribution to $[z^q]$, getting

$$[z^q] (-1)^q \sum_{l=0}^{n} {n\choose l}^2 (1+z)^{2n-2l} \sum_{p\ge 0} {2l\choose p} (-1)^p z^p \\ = [z^q] (-1)^q \sum_{l=0}^{n} {n\choose l}^2 (1+z)^{2n-2l} (1-z)^{2l} \\ = [z^q] (-1)^q [w^n] (1+w(1-z)^2)^n (1+w(1+z)^2)^n \\ = [z^q] [w^n] (1+w(1-z)^2)^n (1+w(1+z)^2)^n.$$

Re-write this as

$$[z^q] [w^n] ((w(1+z^2)+1)^2 - 4 w^2 z^2)^n \\ = [z^q] [w^n] \sum_{p=0}^n {n\choose p} (-1)^p 2^{2p} w^{2p} z^{2p} (w(1+z^2)+1)^{2n-2p} \\ = [z^q] \sum_{p=0}^n {n\choose p} (-1)^p 2^{2p} z^{2p} [w^{n-2p}] (w(1+z^2)+1)^{2n-2p} \\ = [z^q] \sum_{p=0}^n {n\choose p} (-1)^p 2^{2p} z^{2p} {2n-2p\choose n-2p} (1+z^2)^{n-2p}.$$

We observe at this point that we get zero here when $q$ is odd, which agrees with the target formula. We are thus justified in putting $q=2l$ to get

$$[z^l] \sum_{p=0}^n {n\choose p} (-1)^p 2^{2p} z^{p} {2n-2p\choose n-2p} (1+z)^{n-2p} \\ = \sum_{p=0}^n {n\choose p} (-1)^p 2^{2p} {2n-2p\choose n-2p} {n-2p\choose l-p}.$$

Note that

$${n\choose p} {2n-2p\choose n-2p} {n-2p\choose l-p} = \frac{(2n-2p)!}{p! \times (n-p)! \times (l-p)! \times (n-l-p)!} \\ = {l\choose p} \frac{(2n-2p)!}{(n-p)! \times l! \times (n-l-p)!} = {l\choose p} {2n-2p\choose n-p} {n-p\choose l}.$$

Re-indexing we get for the sum

$$(-1)^n 2^{2n} \sum_{p=0}^n {l\choose n-p} {2p\choose p} {p\choose l} (-1)^p 2^{-2p} \\ = (-1)^n 2^{2n} \sum_{p=0}^n {2p\choose p} (-1)^p 2^{-2p} [z^{n-p}] (1+z)^l [w^l] (1+w)^p \\ = (-1)^n 2^{2n} [z^n] (1+z)^l [w^l] \sum_{p=0}^n {2p\choose p} (-1)^p 2^{-2p} z^p (1+w)^p.$$

We may once more extend $p$ to infinity because there is no contribution from the sum term to the coefficient extractor $[z^n]$ when $p\gt n,$ obtaining

$$(-1)^n 2^{2n} [z^n] (1+z)^l [w^l] \sum_{p\ge 0} {2p\choose p} (-1)^p 2^{-2p} z^p (1+w)^p \\ = (-1)^n 2^{2n} [z^n] (1+z)^l [w^l] \frac{1}{\sqrt{1+z(1+w)}} \\ = (-1)^n 2^{2n} [z^n] (1+z)^l [w^l] \frac{1}{\sqrt{1+z+wz}} \\ = (-1)^n 2^{2n} [z^n] (1+z)^{l-1/2} [w^l] \frac{1}{\sqrt{1+wz/(1+z)}} \\ = (-1)^n 2^{2n} [z^n] (1+z)^{l-1/2} {2l\choose l} (-1)^l 2^{-2l} z^l \frac{1}{(1+z)^l} \\ = (-1)^{n-l} 2^{2n-2l} {2l\choose l} [z^{n-l}] \frac{1}{\sqrt{1+z}} \\ = (-1)^{n-l} 2^{2n-2l} {2l\choose l} {2n-2l\choose n-l} (-1)^{n-l} 2^{-(2n-2l)} \\ = {2l\choose l} {2n-2l\choose n-l}.$$

This is the claim. Credit goes to the Egorychev method which was presented here in formal power series notation.