3. Given the function $f(x)$ satisfies:
$$
\begin{array}{l}
f(1)=\frac{1}{4}, \\
4 f(x) f(y)=f(x+y)+f(x-y)(x, y \in \mathbf{R}) .
\end{array}
$$

Then $f(2019)=$.
(A) $\frac{1}{2}$
(B) $-\frac{1}{2}$
(C) $\frac{1}{4}$
(D) $-\frac{1}{4}$
3. B.

Let $x=1, y=0$, we get $f(0)=\frac{1}{2}$.
Let $x=1, y=1$, we get
$$
4 f^{2}(1)=f(2)+f(0) \text {. }
$$

Thus, $f(2)=-\frac{1}{4}$.
Let $x=2, y=1$, we get
$$
4 f(1) f(2)=f(3)+f(1) \text {. }
$$

Thus, $f(3)=-\frac{1}{2}$.
Let $x=n, y=1$, we get
$$
f(n)=f(n+1)+f(n-1) \text {. }
$$

Similarly, $f(n+1)=f(n+2)+f(n)$.
Combining these, we get $f(n+2)=-f(n-1)$.
Thus, $f(n+6)=f(n)$.
Therefore, the period of $f(x)$ is 6.
Hence, $f(2019)=f(336 \times 6+3)=f(3)=-\frac{1}{2}$.