10. Clearly, $f(x) \leqslant 6 x$.

Then $6 x-f(x)=f(f(x)) \leqslant 6 f(x)$
$$
\begin{array}{l}
\Rightarrow f(x) \geqslant \frac{6}{7} x \\
\Rightarrow 6 x-f(x)=f(f(x)) \geqslant \frac{6}{7} f(x) \\
\Rightarrow f(x) \leqslant \frac{6 x}{1+\frac{6}{7}} \\
\Rightarrow 6 x-f(x)=f(f(x)) \leqslant \frac{6 f(x)}{1+\frac{6}{7}} \\
\Rightarrow f(x) \geqslant \frac{6 x}{1+\frac{6}{1+\frac{6}{7}}} .
\end{array}
$$

Let $a_{n}=\frac{6}{1+a_{n-1}}$. Then
$$
\begin{array}{l}
a_{1}=6, \text { and } \\
a_{2} x \leqslant f(x) \leqslant a_{1} x, \\
a_{4} x \leqslant f(x) \leqslant a_{3} x, \\
\cdots \cdots \\
a_{2 n} x \leqslant f(x) \leqslant a_{2 n-1} x .
\end{array}
$$

It is easy to prove that $\left\{a_{2 n-1}\right\}$ is monotonically decreasing, $\left\{a_{n}\right\}$ is monotonically increasing, and
$$
a_{2 n}<2<a_{2 n-1} \text {. }
$$

Let $\lim _{n \rightarrow \infty} a_{2 n-1}=s, \lim _{n \rightarrow \infty} a_{2 n}=t$. Then
$$
\begin{array}{l}
s=\frac{6}{1+t}, t=\frac{6}{1+s} \\
\Rightarrow s=t=2 \\
\Rightarrow f(x)=2 x .
\end{array}
$$
10. (20 points) Find the function $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$, such that $f(f(x))=6 x-f(x)$.