Given $0 \leqslant x, y, z \leqslant 1$, solve the equation:
$$\frac{x}{1+y+z x}+\frac{y}{1+z+x y}+\frac{z}{1+x+y z}=\frac{3}{x+y+z} .$$
3. It is not difficult to prove: $\frac{x}{1+y+z x} \leqslant \frac{1}{x+y+z}$, $\frac{y}{1+z+x y} \leqslant \frac{1}{x+y+z}$, $\frac{z}{1+x+y z} \leqslant \frac{1}{x+y+z}$. Therefore, if the equality holds, it is easy to get $x=y=z=1$.

Note: This problem can also be solved as follows: First prove $\frac{x}{1+y+z x} \leqslant \frac{x}{x+y+z}$, etc., so we can get $x+y+z \geqslant 3$, hence $x=y=z=1$.