Let $f(x, y)\equiv(y-x)\left(\frac{y-1}{y}\right)^{1-x}$, where $0<x<1, y>\hat{y}(x)$. Here $\hat{y}(x)\equiv\arg max_y [f(x, y)-x]$ is the unique peak of $f(x, y)$ as a function of $y$.
The implicit function $y=y(x)$ is given by $$f(x, y)=k\cdot f(x, y-\frac{k-1}{k}f(x, y)),$$ where $k>1$ is a constant.
I'd like to show $y'(x)<0$ for all $0<x<1$. Is there anyone can help?
My work so far: let $z\equiv y-\frac{k-1}{k}f(x, y)$. Then $z'(x)=y'(x)\left[1-\frac{k-1}{k}f_y(x,y)\right]-\frac{k-1}{k}f_x(x,y)$. Using the implicit function theorem, I get $$y'(x)\cdot\left[k f_y(x, z)(1-f_y(x,y))+f_y(x, y)(f_y(x, z)-1)\right]=f_x(x, y)\left[1+(k-1)f_y(x, z)\right]-k f_x(x, z)$$.
We can show $z<\hat{y}(x)<y$, thus $f_y(x, y)<1$ and $f_y(x, z)>1$. So $\left[k f_y(x, z)(1-f_y(x,y))+f_y(x, y)(f_y(x, z)-1)\right]>0$. It follows that the sign of $y'(x)$ is the same as that of $$f_x(x, y)\left[1+(k-1)f_y(x, z)\right]-k f_x(x, z)$$.
The above expression is the part that I have a difficulty in determining the sign. Can anyone help?
