Induction Proof for $F_{2n} = F^2_{n+1} - F^2_{n-1}$

You're given $$f_{2n-1}=f_n^2+f_{n-1}^2\tag1$$ Replace $n$ with $n+1$ to get $$f_{2n+1}=f_{n+1}^2+f_n^2\tag2$$ Subtract (1) from (2) to get $$f_{2n+1}-f_{2n-1}=f_{n+1}^2-f_{n-1}^2\tag3$$ But the left side of (3) is $f_{2n}$.

Method 1. By induction on $n$ we have $F_n=(a^n-b^n)/\sqrt 5$ where $a=(1+\sqrt 5)/2$ and $b=(1-\sqrt 5)/2=-1/a.$ Plug this into your formula.

Method 2. Let $M$ be the $2\times2$ matrix with top row $(1,1)$ and bottom row $(1,0).$ By induction on $n,$ the top row of $M^n$ is $(F_{n+1},F_n)$ and the bottom row of $M^n$ is $(F_n,F_{n-1}).$ Consider that $M^{2 n} =(M^n)^2$.

Method 3. Your method. For brevity let $$F_{n-1}=a, \quad F_n=b,\quad F_{n+1}=c,\quad F_{n+2}=d.$$ From your 3rd line and the inductive hypothesis, and given $F_{2 n-1}=F_n^2+F_{n-1}^2$, we have $$F_{2 n+2}=2 F_{2 n}+F_{2 n-1}=2 (c^2- a^2) +(b^2+a^2)=$$ $$=2 c^2+b^2-a^2 =2 (d-b)^2+b^2-(c-b)^2=$$ $$=2(d-b)^2 +b^2-((d-b)-b)^2=d^2-b^2=F_{n+2}^2-F_n^2.$$

As this is a difficult result to prove, I shall prove another result and then derive the said result through it.

Proposition: $F_{n+m} = F_n F_{m-1} + F_{n+1} F_m$

Base cases: $n = 1$:

$\begin{align} F_{m+1} &= F_m + F_{m-1} \\ &= F_2 F_m + F_1 F_{m-1} \end{align}$

Case $n = 2$:

$\begin{align} F_{m+2} &= F_{m+1} + F_m \\ &= F_{m-1} + 2 F_m \\ &= F_2 F_{m-1} + F_3 F_m \end{align}$

Hypothesis: $F_{k+m-1} = F_{k-1} F_{m-1} + F_k F_m$ and $F_{k + m} = F_k F_{m-1} + F_{k +1} F_m$

Induction step: Adding both the hypothesis statements

$\begin{align} F_{k + m + 1} &= F_{k+m} + F_{k+m-1} \\ &=F_{k+1} F_{m-1} + F_{k+1+1} F_m \end{align}$

Hence, proved.

$F_{n+m} = F_n F_{m-1} + F_{n+1} F_m$

Putting $n = m$:

$\begin{align} F_{2n} &= F_n F_{n-1} + F_{n+1} F_n \\ &= (F_{n+1} - F_{n-1}) F_{n-1} + F_{n+1} F_n \\ &= F_{n+1} F_{n-1} + F_{n+1} F_n - F^2_{n-1} \\ &= F_{n+1}(F_{n-1} + F_n) - F^2_{n-1} \\ &= F^2_{n+1} - F^2_{n-1} \end{align}$