The formal power series of the Fibonacci sequence is given by $$ \sum_{n \geq 0} f_nq^n=q+q^2+2q^3+3q^4+5q^5+\cdots $$ whereby $f_n=f_{n-1}+f_{n-2}$ denotes the $n$-th Fibonacci number. Its closed form, call it $\Phi(q)$, is given by
$$ \Phi(q)=\frac{q}{1-q-q^2} $$
The function $\Phi(q)$ is rational. We can see this by just calculating the above or by knowing the following theorem.
Theorem: The ordinary generating function of a sequence can be expressed as a rational function if and only if the sequence is a linear recurrence relation.
Now, consider the sequence $a_n=f_n^2$. The generating function of $a_n$ is given by the following formal power series.
$$ \sum_{n \geq 0} f_n^2q^n=f_1^2q + f_2^2q^2+f_3^2q^3+\cdots $$
I want to show that $\Phi'(q)$ is rational without calculating its closed form.
My approach: We can show that the sum of squares of consecutive Fibonacci numbers is a Fibonacci number namely $f_n^2 + f_{n+1}^2=f_{2n+1}=f_{2n}+f_{2n-1}$, thus the sequence $a_n$ is a linear recurrence relation. By applying the above theorem, we are done.
Question: Can we tell from the closed form $\Phi(q)$ of the generating function of the usual Fibonacci sequence that $\Phi'(q)$ is rational? If not, is there a different approach for showing that $\Phi'(q)$ is rational without actually calculating it?