Find $f(7)$ given $f(x)f(y)=f(x+y)+f(x-y)$ and $f(1)=3$ 
Let $f:\Bbb R\to\Bbb R$ such that $f(1)=3$ and $f$ satisfies the functional equation
$$f(x)f(y) = f(x+y) + f(x-y)$$
Find the value of $f(7)$.

Attempt:
If $x=1$ and $y=0$, we find
$$f(1) f(0) = 2f(1) \implies f(0) = 2$$
If we fix $y=1$, we get the recurrence relation
$$\begin{cases} f(0) = 2 \\ f(1) = 3 \\ f(x + 1) - 3f(x) + f(x - 1) = 0 & \text{for } x\ge1 \end{cases}$$
From here I can solve for $f(x)$ or sequentially compute $f(2),f(3),f(4),\ldots$ to arrive at $f(7) = 843$.
Question:

Is there a more elegant way of finding $f(7)$ directly from the functional equation?

 A: Well, is not too fancy, but you could do
$f(7)=f(4+3)=f(4)f(3)-f(1),$
and using the fact that
$f(4)=f(3)f(1)-f(2),$
you could compute $f(7)$ by knowing only $f(1),\;f(2),$ and $f(3)$.
A: Evaluate $\pmb{f(0)}$
$f(1)=3$ and
$$
f(x)f(y)=f(x+y)+f(x-y)\tag1
$$
Setting $x=1$ and $y=0$, $(1)$ gives $f(0)=2$.

Determine and Solve a Recursion
Setting $y=1$, $(1)$ gives
$$
3f(x)=f(x+1)+f(x-1)\tag2
$$
Solving the second order linear recurrence in $(2)$ and using $f(0)=2$ and $f(1)=3$ gives, for $x\in\mathbb{Z}$,
$$
\begin{align}
f(x)&=\left(\frac{3+\sqrt5}2\right)^{\large\!x}+\left(\frac{3-\sqrt5}2\right)^{\large\!x}\tag{3a}\\
&=\phi^{2x}+\phi^{-2x}\tag{3b}\\[6pt]
&=L_{2x}\tag{3c}
\end{align}
$$
where $L_n$ is the $n^\text{th}$ Lucas Number (as Akiva Weinberger mentioned in a comment).

Check the Functional Equation $\pmb{(1)}$
If $(1)$ has a solution, then it is given by $(3)$. However, to show that $(1)$ has a solution, we need to verify that $(3)$ satisfies $(1)$.
$$
\begin{align}
f(x)f(y)
&=\left(\phi^{2x}+\phi^{-2x}\right)\left(\phi^{2y}+\phi^{-2y}\right)\tag{4a}\\[4pt]
&=\phi^{2(x+y)}+\phi^{-2(x+y)}+\phi^{2(x-y)}+\phi^{-2(x-y)}\tag{4b}\\[6pt]
&=f(x+y)+f(x-y)\tag{4c}
\end{align}
$$
So, $(1)$ is satisfied.

Apply to the Question
Thus,
$$
\begin{align}
f(7)
&=L_{14}\tag{5a}\\[4pt]
&=843\tag{5b}
\end{align}
$$
