$$\begin{array}{r|cccccccc}\color{Red}{1}f(x) & \color{Red}{F_0} & + & \color{Red}{F_1}x & + & \color{Red}{F_2}x^2 & + & \color{Red}{F_3}x^3 & \cdots \\
\hline \color{Green}{x}f(x) & & & \color{Green}{F_0}x & + & \color{Green}{F_1}x^2 & + & \color{Green}{F_2}x^3 & \cdots \\
\hline \color{Blue}{x^2}f(x) & & & & & \color{Blue}{F_0}x^2 & + & \color{Blue}{F_1}x^3 & \cdots \\
\hline (\color{Red}{1}-\color{Green}{x}-\color{Blue}{x^2})f(x) & \color{Red}{F_0} & + & (\color{Red}{F_1}-\color{Green}{F_0})x & + & (\color{Red}{F_2}-\color{Green}{F_1}-\color{Blue}{F_0})x^2 & + & (\color{Red}{F_3}-\color{Green}{F_2}-\color{Blue}{F_1})x^3 & \cdots \\\end{array}$$
This is $F_0+(F_1-F_0)x+0+0+\cdots=x$, since $F_{n+2}-F_{n+1}-F_n=0$ for all $n\ge0$. Once you reach the functional equation $(1-x-x^2)F(x)=x$, you divide for the closed-form of $F(x)$.
For the generating functions of $F_{2n}$ and $F_{2n+1}$, you could perhaps use identities like
$$F_{2n}=\sum_{k=1}^nF_{2k-1} \qquad F_{2n+1}=1+\sum_{k=1}^nF_{2k}.$$
However, in my opinion, for all three GFs it would be easier to just use Binet's formula
$$F_n=\frac{\varphi^n-\bar{\varphi}^n}{\varphi~-~\bar{\varphi}\,},\qquad \varphi,\bar{\varphi}=\frac{1\pm\sqrt{5}}{2}.$$
One way of proving Binet's formula is with linear algebra, but another way is proving it using the very generating function we just looked at (and using partial fraction decomposition), and in the latter case it is not applicable to the original GF but afterwards can still be used for the other two.