On a function with a (complicated) functional equation. 
Let $g(x,y)$ be a function such that:
I. $-1\lt g(x,y)\lt1.$
II.  $$\ln(\frac{1+g(x,y)}{1-g(x,y)})+2y\tan^{-1}(yg(x,y))=2(y^2+1)x,$$ for $x\in\mathbb R, y\gt1.$
  Then
i. Show that $g(x,y)$ is increasing in $x.$
ii. Find $$\lim_{y\rightarrow\infty} g(x,y).$$
iii. Show that $g(x,y)$ is differentiable.
iv. Find $\lim_{y\rightarrow\infty}\frac{\partial}{\partial x}g(x,y).$  

My question is: how to evaluate the limit maked as ii. ?
I have made an attempt to solve the problem, but I still lack a pice to complete the puzzle.
So it helps a lot if someone can tell me how to evaluate the limit
Thanks in advance.
 A: Since it looks messy to put my efforts in the question body, I decided to write an answer to explain what I have shown and where I got stuck, in an attempt to get an answer. :)  

The overall look:
  I tried to show first that $g$ is differentiable, then that $g(x,y)$ is increasing in $x.$ After this, I tried to find the limit $\lim_{y\rightarrow\infty}g(x,y),$ but to no avail. Finally, if I can find the above limit, then I can also compute $\lim_{y\rightarrow\infty}\frac{\partial}{\partial x}g(x,y).$  

Differentiability of $g$:
Define $f(x,y,z)=\ln(\frac{1+z}{1-z})+2y\tan^{-1}(yz)-2(y^2+1)x.$
Then, for any pair $(x,y)$ with $x\in\mathbb R, y\gt0,$ we find that $\lim_{z\rightarrow1}f(x,y,z)=\infty,$ and $\lim_{z\rightarrow-1}f(x,y,z)=-\infty.$ Hence, by the mean-value theorem, there is $z_0$ such that $-1\lt z_0\lt1$ and $f(x,y,z_0)=0.$ Thus we can apply the implicit-function theorem, and deduce that there is a nbd $U$ of $(x,y),$ and a differentiable function $g_U:U\rightarrow \mathbb R$ such that $g_U$ is the unique solution to the functional equation on $U.$ This shows that $g\mid_U=g_U,$ and hence $g$ is differentiable everywhere, i.e. $g$ is differentiable.  
Monotonicity of $g$ in $x$:
By the implicit-function theorem, we find that $$\frac{\partial}{\partial x}g(x,y)=\frac{2(y^2+1)}{\frac{1}{1+g(x,y)}-\frac{1}{1-g(x,y)}+\frac{2y^2}{1+(yg(x,y))^2}}\gt0.$$
This shows that $g(x,y)$ is increasing in $x.$  
On the limit $\lim_{y\rightarrow\infty}g(x,y)$ : 
I rewrite the equation as:
$$2y(\tan^{-1}(yg(x,y))-xy)=2x-\ln(\frac{1+g(x,y)}{1-g(x,y)})$$
$$\tan^{-1}(yg(x,y))=xy+\frac{x}{y}-\frac{\ln(\frac{1+g(x,y)}{1-g(x,y)})}{2y}$$
$$g(x,y)=\frac{tan(xy+\frac{x}{y}-\frac{\ln(\frac{1+g(x,y)}{1-g(x,y)})}{2y})}{y}.$$
But I still cannot infer what the limit should be from this.  
The Second Limit
is a direct consequence of the implicit-function theorem and the above limit.  
This is all I know about the question, in the dearth of a piece to complete the puzzle.
Hope this helps clarify my question.
A: I posted this question in another community, and got an answer, so I post that answer here, for reasons of completeness.
We distinguish between three cases:
I. $x\gt0$
Divide the whole equation by $y^2,$ and find, as $\operatorname{tan}^{-1}$ is bounded, $\lim_{y\rightarrow\infty}\frac{\operatorname{ln}(\frac{1+g}{1-g})}{y^2}=2x.$ This implies that $\lim_{y\rightarrow\infty}\operatorname{ln}(\frac{1+g}{1-g})=+\infty,$ thus $\lim_{y\rightarrow\infty}g(x,y)=1.$  
II. $x=0$
This time divide the functional equation by $y,$ and find that $\lim_{y\rightarrow\infty}\frac{\operatorname{ln}(\frac{1+g}{1-g})}{y}=0.$ Thus, in view of the last equality in my previous answer, we coclude that $\lim_{y\rightarrow\infty}g(x,y)=0.$  
III. $x\lt0$
Similarly as in I, we find that $\lim_{y\rightarrow\infty}\ln(\frac{1+g}{1-g})=-\infty,$ hence $\lim_{y\rightarrow\infty}g(x,y)=-1.$  
From the above we conclude that the limit in $iv.$ is equal to $\begin{cases}0&x\not=0\\1&x=0\end{cases}.$ (I think we have to apply L'Hôspital twice in the case $x\not=0.$)
Hope this helps. ;P
