Given a smooth function $f(x,y)$, find $\frac{\partial^2f}{\partial(\ln x)\partial(\ln y)}$.
The solution is not hard
$$\frac{\partial^2f}{\partial(\ln x)\partial(\ln y)}=\partial_{\ln x}\frac{\partial f}{1/y\partial y}$$
$$=\partial_{\ln x}(yf_y)$$ $$=y\partial_{\ln x}(f_y)+f_y\frac{\partial y}{\partial {\ln x}}$$ $$=yx\partial_xf_y-f_y\frac{f_x}{1/xf_y}$$ $$=xyf_{xy}-xf_x$$
Similarly, we can find:
$$\frac{\partial^2f}{\partial(\ln y)\partial(\ln x)}=xyf_{xy}-yf_y$$
Obviously $$\frac{\partial^2f}{\partial(\ln y)\partial(\ln x)}=\frac{\partial^2f}{\partial(\ln x)\partial(\ln y)}$$
So
$$xf_x=yf_y$$
What's wrong with it?
Because $\frac{\partial y }{\partial \ln x} = \frac{\partial x}{\partial \ln y} = 0$ ($y$ is treated as a constant when differentiating WRT $x$ and vice versa). So the answer in both cases is just $xyf_{xy}$.