Using the multinomial theorem, one can show that $f(n)$ is the coefficient of $x^n$ of the polynomial $(1+x+x^2)^n$. There are $3$ obvious ways to show the equation in the title:
First, you can square the $2$ sides of the equation and then multiply with $1-2x-3x^2$. In that case, the coefficients of the left formal power series seems too much.
Second, we can factorize the polynomial, $$ 1-2x-3x^2=(-3)(x+1)(x-1/3)=(1+x)(1-3x) $$
$$\mbox{and use the formula ,}\quad \sqrt{1+F(x)}=\sum_{n\geq 0}^{}(-1)^n \frac{1}{4^n}{2n \choose n}F(x)^n $$ for $F(x)=x , G(x)=-3x$, then multiply the formal power series and see if the equation holds.
The third way is to use the previous method for $F(x)=-(2x+3x^2)$. In the 2 last methods the numbers arent very far from those we want, but i cant prove it. I think that i am missing some identity with binomial coefficients and thats why i cant solve it.
Here's a way that doesn't require knowing the result ahead of time. \begin{align} \sum_{n \ge 0} f(n) x^n &= \sum_{n \ge 0}\sum_{k=0}^{\left\lfloor n/2\right\rfloor} \binom{2k}{k}\binom{n}{2k} x^n \\ &= \sum_{k \ge 0}\binom{2k}{k} \sum_{n\ge 2k} \binom{n}{2k} x^n \\ &= \sum_{k \ge 0}\binom{2k}{k} \frac{x^{2k}}{(1-x)^{2k+1}} \\ &= \frac{1}{1-x}\sum_{k \ge 0}\binom{2k}{k} \left(\left(\frac{x}{1-x}\right)^2\right)^k \\ &= \frac{1}{1-x}\cdot\frac{1}{\sqrt{1-4\left(\frac{x}{1-x}\right)^2}} \\ &= \frac{1}{\sqrt{(1-x)^2-4x^2}} \\ &= \frac{1}{\sqrt{1-2x-3x^2}} \end{align}