12．已知定义在$\mathbf{R}$上且不恒为$0$的函数$f(x)$，若对任意的$x$，$y \in \mathbf{R}$，都有$f(xy) = xf(y) + yf(x)$，则（ ）
A．函数$f(x)$是奇函数
B．对$\forall n \in \mathbf{N}^*$，有$f(x^n) = nf(x)$
C．若$f(2) = 2$，则$f(2) + f(2^2) + f(2^3) + \cdots + f(2^n) = (n + 1)2^n - 2$
D．若$f(2) = 2$，则$\frac{f\left( \frac{1}{2} \right)}{1} + \frac{f\left( \left( \frac{1}{2} \right)^2 \right)}{2} + \frac{f\left( \left( \frac{1}{2} \right)^3 \right)}{3} + \cdots + \frac{f\left( \left( \frac{1}{2} \right)^{10} \right)}{10} = - \frac{1023}{1024}$
解 对于选项A
令$x = y = 1$，则$f(1) = f(1) + f(1)$，$\therefore f(1) = 0$；
令$x = y = - 1$，则$f(1) = - f( - 1) - f( - 1)$，$\therefore f( - 1) = 0$；
令$y = - 1$，则$f( - x) = - f( - 1) - f(x)$，$\therefore f( - x) = - f(x)$，故A正确．
对于选项B
令$y = x^{n - 1}$，则$f(x^n) = xf(x^{n - 1}) + x^{n - 1}f(x)$，
由$f(x^2) = 2xf(x) \neq 2f(x)$，故B错误．
对于选项C
令$x = 2$，$y = 2$，则$f(4) = 2f(2) + 2f(2) = 8$；
$f(8) = f(2 \cdot 4) = 2f(4) + 4f(2) = 24$，
当$n = 3$时等式左端为$f(2) + f(4) + f(8) = 2 + 8 + 24 = 34$，右端为$(3 + 1)2^3 - 2 = 30$，故C错误．
对于选项D
令$x = 2$，$y = \frac{1}{2}$，则$f(1) = 2f\left( \frac{1}{2} \right) + \frac{1}{2}f(2)$，$\therefore f\left( \frac{1}{2} \right) = - \frac{1}{2}$；
由$f(x^2) = 2xf(x)$，$f(x^3) = 3x^2f(x)$，$f(x^4) = 4x^3f(x)$，归纳得$f(x^n) = nx^{n - 1}f(x)$，
$\therefore f\left( \left( \frac{1}{2} \right)^n \right) = n\left( \frac{1}{2} \right)^{n - 1}f\left( \frac{1}{2} \right) = - n \cdot \left( \frac{1}{2} \right)^n$，
$\therefore \frac{f\left( \left( \frac{1}{2} \right)^1 \right)}{1} + \frac{f\left( \left( \frac{1}{2} \right)^2 \right)}{2} + \cdots + \frac{f\left( \left( \frac{1}{2} \right)^{10} \right)}{10} = - \left( \left( \frac{1}{2} \right)^1 + \left( \frac{1}{2} \right)^2 + \cdots + \left( \frac{1}{2} \right)^{10} \right) = - \frac{1023}{1024}$，故D正确．