Proving this $F_{n+1} \cdot F_{n-1} - F^2_n = (-1)^n$ by induction Where $n \in \mathbb{N}$ 
and
$$
F_n = \begin{cases}
0                 & \text{ if } n = 0 \\
1                 & \text{ if } n = 1 \\
F_{n-1} + F_{n-2} & \text{ if } n > 1
\end{cases}
$$
This is basically describing the famous Fibonacci sequence.
If we try the base case with 1, it works (for $0$ I am not sure...)
When $n = 1$
$2 \cdot0 - 1^2 = (-1)^1 = -1$
For the inductive hypothesis, we assume this $F_{n+1} \cdot F_{n-1} - F^2_n = (-1)^n$ is true.
Now, for the inductive step, we try to prove for $n+1$, so for  $F_{n+2} \cdot F_{n} - F^2_{n+1} = (-1)^{n+1}$.
Since $n$ is always a natural number, and it will be always or even or odd, the $-1$ raised to $n$ will be always either $-1$ (when $n$ is odd) or $1$ (when $n$ is even). 
Thus, $F_{n+1} \cdot F_{n-1} - F^2_n$ = -($F_{n+2} \cdot F_{n} - F^2_{n+1}$).
Or simply:
$$
(-1)^n = -(-1)^{n+1}
$$
$$
1 \cdot (-1)^n = -(-1)\cdot(-1)^n
$$
$$
1 = -(-1)
$$
Which is true.
I don't know if this is sufficient... I can arrive at the second step and say that we know this $(-1)^n$ is equals to $F_{n+1} \cdot F_{n-1} - F^2_n$, but I am not sure...

Your proof is completely incorrect. You need to prove why $$F_{n+2} F_{n} - F_{n+1}^2 = -(F_{n+1} F_{n-1} - F_{n}^2)$$ to prove that $F_{n+2} F_{n} - F_{n+1}^2 = (-1)^{n+1}$.
You can check the base case. For the inductive step, we have
\begin{align}
F_{n+2} F_{n} - F_{n+1}^2 & = (F_{n+1}+F_{n})F_{n} - F_{n+1}^2 & \left(\because F_{n+2} = F_{n+1} + F_{n} \right)\\
& = F_{n+1} F_{n} + F_{n}^2 - F_{n+1}^2\\
& = F_{n+1} F_{n} - F_{n+1}^2 + F_{n}^2\\
& = F_{n+1}(F_{n}-F_{n+1}) + F_{n}^2\\
& = -F_{n+1} F_{n-1} + F_{n}^2 & \left(\because F_{n+1} = F_{n} + F_{n-1}\right)\\
& = -\left(F_{n+1}F_{n-1} - F_{n}^2\right)
\end{align}
Now you are done.
