The number of compositions of $n$ with exactly $k$ parts is $\dbinom{n-1}{k-1}$, so the generating function for the number of compositions with an even number of parts is
$g(x)=\sum_{n\ge 0}\left(\sum_{k\ge 0}\binom{n-1}{2k-1}\right)x^n\;.\tag{1}$
$\displaystyle\sum_{k\ge 0}\binom{n-1}{2k-1}$ is simply the number of subsets of $\{1,\dots,n-1\}$ with an odd number of elements. For $n\le 1$ that’s clearly $0$, so we can rewrite $(1)$ as
$g(x)=\sum_{n\ge 2}\left(\sum_{k\ge 0}\binom{n-1}{2k-1}\right)x^n=x^2\sum_{n\ge 2}\left(\sum_{k\ge 0}\binom{n-1}{2k-1}\right)x^{n-2}=x^2\sum_{n\ge 0}\left(\sum_{k\ge 0}\binom{n+1}{2k-1}\right)x^n\;.$
Now $\displaystyle\sum_{k\ge 0}\binom{n+1}{2k-1}$ is the number of subsets of $\{1,\dots,n+1\}$ having an odd number of elements, and since $n+1\ge 1$, this has a simple closed form that you should know. Let’s say that that closed form is $f(n)$. Then you have
$g(x)=x^2\sum_{n\ge 0}f(n)x^n\;,$
where you should be able to recognize the generating function for $\displaystyle\sum_{n\ge 0}f(n)x^n$ fairly easily.
Added: If you follow the convention that $0$ has one composition, of size $0$, then $g(x)$ should have a constant term $1$ in addition to the terms given by $(1)$.