Problem:
Let $f(x,y)=x-\ln x - y +\ln y$ for $x,y>0$. Prove that there exists a $\delta>0$ and a function $\varphi : (-\delta, \delta)\to \mathbb{R}:\ \varphi \in C^1,\ \varphi(0)<0,\ \forall_{x\in (1-\delta, 1+\delta)}\ f(x,1+(x-1)\varphi(x-1))=0$
I came up with a bit of a solution but there must be an easier way. Also, I'd like find out if mine is correct and how to finish it.
Ok, it is easy to see that IFT won't work for the function $f$ because its derivative is zero on the line $y=x$. My idea is to modify the function.
Let $g(x) = x-\ln x$ and $F(x,y)=\frac{g(x)-g(y)}{x-y}$. We have $F=0 \Rightarrow f=0$. By Lagrange Theorem $F(x,y)=g'(c)$ where $c\in [x,y]$, thus by taking limit with $(x,y)\to (a,a)$ we have $F(a,a)=g'(a) = 1-\frac{1}{a}$. Hence $F$ is continuous on $\mathbb{R}^2_+$.
Now let's show that it's $C^1$ class. We only need to show that derivative is continuous on the line $x=y$. We have \begin{eqnarray*} &\frac{\partial F}{\partial x}(x,y) = \frac{(x-y)g'(x) -(g(x)-g(y))}{(x-y)^2} \end{eqnarray*} Thus \begin{eqnarray*} &\lim _{h\to 0}\frac{\partial F}{\partial x}(x+h,x) = \lim_{h\to 0} \frac{hg'(x+h) -(g(x+h)-g(x))}{h^2} =\\ &\lim_{h\to 0} \frac{hg'(x) + h^2g''(x)- hg'(x) - \frac{1}{2}h^2g''(x) +o(h^2)}{h^2}=\frac{1}{2}g''(x)= \frac{1}{2x^2} \end{eqnarray*} We can do the same for $\frac{\partial F}{\partial y} \Rightarrow$ F has continuous paritial derivatives, hence is of class $C^1$. Moreover $DF(x,y)=(0,0) \Rightarrow \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}=0 \Rightarrow \frac{g'(x)-g'(y)}{x-y}=0 \Rightarrow x=y$ but $DF(x,x)\neq (0,0)$ for $x>0$ (which was proven above). So, $DF(x,y)\neq (0,0)$ for all $(x,y)$.
Hence we can apply IFT: there exists a $\delta >0$ and $C^1$ class function $h:(1-\delta,1+ \delta):\ F(1+x,h(1+x))=0 \Rightarrow f(1+x,h(1+x))=0$. Now we can define $\varphi(x) := \frac{h(x+1)-1}{x}$. All we need o do is to check that $\varphi$ satisfies task conditions. We can write Taylor series for $h$ around the point $(x,y)=(1,1)$ (since we know that $DF(1,1)=(\frac{1}{2},\frac{1}{2})$) and obtain: $\frac{h(x+1)-1}{x} = \frac{1+2x+o(x)-1}{x} \to 2$ as $x\to 0$.
Here the trouble begins: $\varphi$ doesn't satisfy the conditions but it would suffice to take $\bar{F}=-F(x,y)$ instead of $F(x,y)$ and $\bar{h}:\bar{F}(x,\bar{h}(x))=0$, then $\phi := \frac{\bar{h}(x+1)-1}{x} \to -2$ as $x\to 0$ so we can handle this problem. But how to show that $\varphi$ has continuous derivative at $x=0$? Do we have to compute the second Taylor polynomial for $h$? It's possible but seems like a lot of calculations..
I came across this problem on my calculus exam two days ago and it was only a part of a bigger one, so I don't believe the answer is that convoluted. It is the only thing I couldn't figure out so I would appreciate any help.
EDIT: BTW: It should be true for any locally non-constant $C^1$ function $f$, that around it's local maxima/minima there exists a non-identity $C^1$ function $g$ that $f(a+x)=f(a+xg(x))$ for $x$ from neighbourhood of that maxima/minima - $a$. So the task is only a special case of this theorem, am I right?