I assume that you define the Fibonacci function as the natural continuation of the Fibonacci sequence,
$$
F(x) = \frac{\alpha^x-\beta^x}{\sqrt5}
$$
where $\alpha = (1+\sqrt5)/2, \beta=(1-\sqrt5)/2$. If not, I'll just delete this answer and will have wasted my time.
For later use, note that $\alpha\beta+1=0,\alpha+1/\alpha=\sqrt5\text{, and }\beta+1/\beta=-\sqrt5.$
Now notice that for any integer $n$ we have $F(n)=f_n,$ so you wish to prove
$$
F(x+n)=F(n)F(x+1)+F(n-1)F(x)
$$
Using the supposed definition of $F(x)$ the right side of your equality, $F(n)F(x+1)+F(n-1)F(x)$, becomes
$$
\begin{align*}
&= \frac{1}{5}[(\alpha^n-\beta^n)(\alpha^{x+1}-\beta^{x+1})+(\alpha^{n-1}-\beta^{n-1})(\alpha^x-\beta^x)]\\
&= \frac{1}{5}[\alpha^{x+n+1}-\alpha^n\beta^{x+1}-\alpha^{x+1}\beta^n+\beta^{x+n+1}+\alpha^{x+n-1}-\alpha^{n-1}\beta^x-\alpha^x\beta^{n-1}+\beta^{x+n-1}]\\
&=\frac{1}{5}[\alpha^{x+n}(\alpha+1/\alpha)+\beta^{x+n}(\beta+1/\beta)-\alpha^{n-1}\beta^x(\alpha\beta+1)-\alpha^x\beta^{n-1}(\alpha\beta+1)]\\
&=\frac{1}{5}[\alpha^{x+n}(\alpha+1/\alpha)+\beta^{x+n}(\beta+1/\beta)]\quad\text{(using our first observation above)}\\
&=\frac{1}{5}[\alpha^{x+n}\sqrt5-\beta^{x+n}\sqrt5]\quad\text{(using our second and third observations)}\\
&=\frac{\alpha^{x+n}-\beta^{x+n}}{\sqrt5}\\
&=F(x+n)
\end{align*}
$$