5. Let $R$ be the set of all real numbers. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y$, we have
$$
\begin{array}{c}
f(x+f(x+y))+f(x y) \\
=x+f(x+y)+y f(x) .
\end{array}
$$
Below, we discuss in two scenarios.
【Scenario 1】 $f(0) \neq 0$.
Consider $P(0, y)$, we have
$$
f(f(y))+f(0)=f(y)+y f(0) \text {. }
$$

If $y_{0}$ is a fixed point of $f$, setting $y=y_{0}$ in the above equation, we get $y_{0}=1$.
Thus, $x+f(x+1)=1$.
Therefore, $f(x)=2-x$ for all real numbers $x$.
It is easy to verify that $f(x)=2-x$ satisfies the conditions.
【Scenario 2】 $f(0)=0$.
Consider $P(x+1,0)$ and $P(1, y)$, we have
$$
\begin{array}{l}
f(x+f(x+1)+1)=x+f(x+1)+1, \\
f(1+f(y+1))+f(y) \\
=1+f(y+1)+y f(1) .
\end{array}
$$

Setting $x=-1$ in equation (1), we get $f(-1)=-1$.
Then, setting $y=-1$ in equation (3), we get $f(1)=1$.
Thus, equation (3) can be rewritten as
$$
\begin{array}{l}
f(1+f(y+1))+f(y) \\
=1+f(y+1)+y .
\end{array}
$$

If $y_{0}$ and $y_{0}+1$ are both fixed points of $f$, setting $y=y_{0}$ in equation (4), we know that $y_{0}+2$ is also a fixed point of $f$.

Thus, from equations (1) and (2), we know that for any real number $x$, $x+f(x+1)+2$ is a fixed point of $f$, i.e.,
$$
f(x+f(x+1)+2)=x+f(x+1)+2 \text {. }
$$

Substituting $x$ with $x-2$ in the above equation, we get
$$
f(x+f(x-1))=x+f(x-1) \text {. }
$$

Consider $P(x,-1)$, we have
$$
f(x+f(x-1))=x+f(x-1)-f(x)-f(-x) \text {. }
$$

From the above two equations, we know that $f(-x)=-f(x)$, i.e., $f$ is an odd function.
Consider $P(-1,-y)$, and using $f(-1)=-1$, we have
$$
\begin{array}{l}
f(-1+f(-y-1))+f(y) \\
=-1+f(-y-1)+y .
\end{array}
$$

Since $f$ is an odd function, the above equation can be rewritten as
$$
\begin{array}{l}
-f(1+f(y+1))+f(y) \\
=-1-f(y+1)+y .
\end{array}
$$

Adding the above equation to equation (4), we know that $f(y)=y$ for all real numbers $y$.
It is easy to verify that $f(x)=x$ satisfies the conditions.
5. Denote the equation in the problem as $P(x, y)$.
Let $f$ be a function that satisfies the conditions.
Consider $P(x, 1)$, we have
$$
f(x+f(x+1))=x+f(x+1) \text {. }
$$

Thus, for any real number $x$, $x+f(x+1)$ is a fixed point of $f$.

In summary, the functions that satisfy the conditions are
$$
f(x)=x \text { and } f(x)=2-x \text {. }
$$