11 Find all functions $f, g: \mathbf{R} \rightarrow \mathbf{R}$, such that $f(x+y g(x))=g(x)+$ $x f(y), \forall x, y \in \mathbf{R}$.
1. $f(x) \equiv 0, g(x) \equiv 0$ is the trivial solution.
2. Now consider the non-trivial solutions. Let $x=0$, then $f(y g(0))=g(0)$.
   If $g(0) \neq 0$, then $f(x)=g(0)$, i.e., $f(x)=$ constant $c$; thus $g(0)=$ $g(x)+x g(0), g(x)=(1-x) g(0)$. Hence $f(x)=c, g(x)=(1-x) c$ is a solution.
3. If $g(0)=0$, then $f(0)=0$. Let $y=0, f(x)=g(x)+x f(0)=g(x), \forall x \in \mathbf{R}$, so the problem reduces to solving the functional equation $f(x+y f(x))=f(x)+x f(y) (\forall x, y \in \mathbf{R})$. Let $y=0$, then $f(x)=f(x)+x f(0) \Rightarrow f(0)=0$; if $f(x)=0$, then $0=x f(y) \Rightarrow$ $x=0$, i.e.,
   $$
   f(x)=0 \Leftrightarrow x=0 .
   $$

   Let $x=1$, we get
   $$
   f(1+y f(1))=f(1)+f(y),
   $$

   If $f(1) \neq 1$, take $y=\frac{1}{1-f(1)}$, substitute into equation (1), we get $f(1)=0$, which contradicts ( * ). Therefore,
   $$
   \begin{array}{c}
   f(1)=1, \\
   f(1+y)=1+f(y), \quad \forall y \in \mathbf{R} .
   \end{array}
   $$

   In particular, $f(n)=n, n \in \mathbf{Z}$.
   Take $x=n \in \mathbf{Z}, y=z-1$, substitute into the equation, we get
   $$
   \begin{aligned}
   f(n z) & =f(n+(z-1) f(n))=n+n f(z-1) \\
   & =n f(z), \forall n \in \mathbf{Z}, z \in \mathbf{R}, \\
   & \Rightarrow f(r z)=r f(z), \quad \forall r \in \mathbf{Q}, z \in \mathbf{R} .
   \end{aligned}
   $$

   From (3) $f(a)+f(-a)=0=f(a-a)$.
   When $a+b \neq 0$,
   $$
   \begin{array}{r}
   f(a)+f(b)=f\left(\frac{a+b}{2}+\frac{\frac{a-b}{2}}{f\left(\frac{a+b}{2}\right)} f\left(\frac{a+b}{2}\right)\right)+ \\
   f\left(\frac{a+b}{2}+\frac{\frac{b-a}{2}}{f\left(\frac{a+b}{2}\right)} f\left(\frac{a+b}{2}\right)\right) \\
   =f\left(\frac{a+b}{2}\right)+\frac{a+b}{2} f\left(\frac{\frac{a-b}{2}}{f\left(\frac{a+b}{2}\right)}\right)+ \\
   f\left(\frac{a+b}{2}\right)+\frac{a+b}{2} f\left(\frac{\frac{b-a}{2}}{f\left(\frac{a+b}{2}\right)}\right) \\
   =2 f\left(\frac{a+b}{2}\right)=f(a+b),
   \end{array}
   $$

   Therefore, $f(x+y f(x))=f(x)+f(y f(x)), x+y f(x) \neq 0$.
   Also, $f(x+y f(x))=f(x)+x f(y)$,

   Therefore,
   $$
   f(y f(x))=x f(y) .
   $$

   Let $y=1$, then $f(f(x))=x$, so $f$ is bijective.
   Substitute $x$ with $f(x)$, we get $f(x y)=f(x) f(y), \forall x, y \in \mathbf{R}$.
   Let $y=x$, then $f\left(x^{2}\right)=f^{2}(x)>0$; let $y=-x$, then $f\left(-x^{2}\right)=$ $f(x) f(-x)$. This implies $f(x)>0 \Leftrightarrow x>0$. Also, since $f(x-f(x))=$ $f(x)-x=-(x-f(x))$, we have $f(x)-x=0, x \in \mathbf{R}$. Hence
   $$
   f(x)=g(x)=x, \forall x \in \mathbf{R} .
   $$