Three, (20 points) Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for any $x, y \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(4 x-y-1)+4 f(x+y)-16 f(x) \\
=(6 x-2)^{2}+45 y^{2}
\end{array}
$$
Three, taking $y=0$ in equation (1), we get
$$
f(4 x-1)-12 f(x)=(6 x-2)^{2} \text {. }
$$

Taking $y=3 x-1$ in equation (1), we get
$$
\begin{array}{l}
4 f(4 x-1)-15 f(x) \\
=(6 x-2)^{2}+45(3 x-1)^{2} .
\end{array}
$$
(3) $-4 \times$ (2) yields $f(x)=(3 x-1)^{2}$.

Upon verification, it is known that $f(x)=(3 x-1)^{2}$ satisfies equation (1).
Note: This problem can also be solved by taking $y=4 x$ and $y=-x-1$ respectively, to find $f(x)$ (first find the values of $f(0)$ and $f(-1)$).