设 $f(u)$ 可导, $z=x y f\left(\frac{y}{x}\right)$, 若 $x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=x y(\ln y-\ln x)$, 则 $(\quad)$
A. $f(1)=\frac{1}{2}, f^{\prime}(1)=0$
B. $f(1)=0, f^{\prime}(1)=\frac{1}{2}$
C. $f(1)=1, f^{\prime}(1)=0$
D. $f(1)=0, f^{\prime}(1)=\frac{1}{2}$
$\frac{\partial z}{\partial x}=y\left[f\left(\frac{y}{x}\right)+x f^{\prime}\left(\frac{y}{x}\right) \cdot-\frac{y}{x^{2}}\right], \frac{\partial z}{\partial y}=x\left[f\left(\frac{y}{x}\right)+y f^{\prime}\left(\frac{y}{x}\right) \cdot \frac{1}{x}\right]$,
则 $x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=2 x y f\left(\frac{y}{x}\right)=x y \ln \left(\frac{y}{x}\right)$. 因此 $2 f\left(\frac{y}{x}\right)=\ln \left(\frac{y}{x}\right)$, 即 $f(u)=\frac{1}{2} \ln u$.
故 $f(1)=0, f^{\prime}(1)=\frac{1}{2}$.

因此答案为：D