$$ {\frac{\frac{1}{u^2}+\frac{2}{u}}{\frac{1}{u}+1}=\frac{\frac{1}{u^2}+\frac{2u}{u^2}}{\frac{1}{u}+\frac{u}{u}}=\frac{\frac{1+2u}{u^2}}{\frac{u+1}{u}}=\frac{1+2u}{u^2}\cdot \frac{u}{u+1}=\frac{2u+1}{u\left( u+1 \right)}，即：} \\ { 2f\left( \frac{1}{u} \right) +\frac{1}{u^2}f\left( u \right) =\frac{2u+1}{u\left( u+1 \right)}\Longleftrightarrow 2f\left( \frac{1}{x} \right) +\frac{1}{x^2}f\left( x \right) =\frac{2x+1}{x\left( x+1 \right)}} \\ { 得到方程组：\begin{cases} 2f\left( x \right) +x^2f\left( \frac{1}{x} \right) =\frac{x^2+2x}{x+1}\\ 2f\left( \frac{1}{x} \right) +\frac{1}{x^2}f\left( x \right) =\frac{2x+1}{x\left( x+1 \right)}\\ \end{cases}，\left( 2 \right) 代入\left( 1 \right) 有：} \\ { 2f\left( \frac{1}{x} \right) =\frac{2x+1}{x\left( x+1 \right)}-\frac{1}{x^2}f\left( x \right) \Longrightarrow f\left( \frac{1}{x} \right) =\frac{2x+1}{2x\left( x+1 \right)}-\frac{1}{2x^2}f\left( x \right) } \\ { x^2f\left( \frac{1}{x} \right) =x^2\left[ \frac{2x+1}{2x\left( x+1 \right)}-\frac{1}{2x^2}f\left( x \right) \right] =\frac{2x^2+x}{2\left( x+1 \right)}-\frac{1}{2}f\left( x \right) } \\ {2f\left( x \right) +x^2f\left( \frac{1}{x} \right) =2f\left( x \right) +\left[ \frac{2x^2+x}{2\left( x+1 \right)}-\frac{1}{2}f\left( x \right) \right] =\frac{3}{2}f\left( x \right) +\frac{2x^2+x}{2\left( x+1 \right)}} \\ {于是：} \\ {\frac{3}{2}f\left( x \right) +\frac{2x^2+x}{2\left( x+1 \right)}=\frac{x^2+2x}{x+1}\Longrightarrow \frac{3}{2}f\left( x \right) =\frac{x^2+2x}{x+1}-\frac{2x^2+x}{2\left( x+1 \right)}} \\ {=\frac{2x^2+4x}{2\left( x+1 \right)}-\frac{2x^2+x}{2\left( x+1 \right)}=\frac{3x}{2\left( x+1 \right)}\Longrightarrow \frac{3}{2}f\left( x \right) =\frac{3x}{2\left( x+1 \right)}} \\ {故f\left( x \right) =\frac{x}{x+1}} \\ { 注：解法2的方程组也可采用下列办法快速运算} \\ {\begin{cases} 2f\left( x \right) +x^2f\left( \frac{1}{x} \right) =\frac{x^2+2x}{x+1}\\ 2f\left( \frac{1}{x} \right) +\frac{1}{x^2}f\left( x \right) =\frac{2x+1}{x\left( x+1 \right)}\\ \end{cases}\Longrightarrow \begin{cases} 2f\left( x \right) +x^2f\left( \frac{1}{x} \right) =\frac{x^2+2x}{x+1}\\ 2x^2f\left( \frac{1}{x} \right) +f\left( x \right) =\frac{2x^2+x}{x+1}\\ \end{cases}} \\ { 用\left( 1 \right) \times 2-\left( 2 \right) 有：} \\ { \left[ 4f\left( x \right) +2x^2f\left( \frac{1}{x} \right) \right] -\left[ 2x^2f\left( \frac{1}{x} \right) +f\left( x \right) \right] =\frac{2x^2+4x}{x+1}-\frac{2x^2+x}{x+1}\Longrightarrow } \\ {3f\left( x \right) =\frac{3x}{x+1}，故f\left( x \right) =\frac{x}{x+1}} $$