3 Property Method
In solving functional equations, the property method refers to first obtaining some properties of the function, such as monotonicity, odd/evenness, periodicity, symmetry, injectivity, surjectivity, etc., through appropriate assignments and substitutions, and then using these properties to find the solution of the functional equation ${ }^{[4]}$.

Example 4 Let the function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfy that for any $x, y \in \mathbf{R}$, we have
$$
f(f(x)+2 y)=6 x+f(f(y)-x) .
$$

Find the function $f(x)$. ${ }^{[6]}$
(2015, Thailand Mathematical Olympiad)
Solve: Substituting $y=-\frac{f(x)}{2}$ into equation (1), we get
$$
f\left(f\left(-\frac{f(x)}{2}\right)-x\right)=f(0)-6 x(x \in \mathbf{R}) \text {. }
$$

Therefore, for $y \in \mathbf{R}$, taking $x=\frac{f(0)-y}{6}$, we have
$$
f\left(f\left(-\frac{1}{2} f\left(\frac{f(0)-y}{6}\right)\right)-\frac{f(0)-y}{6}\right)=y \text {. }
$$

This shows that $f$ is surjective.
Next, we prove that $f$ is injective.
Suppose $x_{1} 、 x_{2} \in \mathbf{R}$, satisfying $f\left(x_{1}\right)=f\left(x_{2}\right)$.
Since $f$ is surjective, there exists $y \in \mathbf{R}$ such that
$$
f(y)=x_{1}+x_{2} \text {. }
$$

Thus, $6 x_{1}+f\left(x_{2}\right)=6 x_{1}+f\left(f(y)-x_{1}\right)$
$$
\begin{array}{l}
=f\left(f\left(x_{1}\right)+2 y\right)=f\left(f\left(x_{2}\right)+2 y\right) \\
=6 x_{2}+f\left(f(y)-x_{2}\right)=6 x_{2}+f\left(x_{1}\right) .
\end{array}
$$

Therefore, $x_{1}=x_{2}$, which means $f$ is injective.
In equation (1), taking $x=0$, we get
$$
f(f(0)+2 y)=f(f(y))
$$
$\Rightarrow f(y)=f(0)+2 y=2 y+a$, where $a=$ $f(0)$ is a real constant.

Upon verification, $f(x)=2 x+a$ is a solution to the original equation.