everywhere differentiable function — an interesting problem

Question: Determine all the functions f, which are everywhere differentiable and satisfy $f(x)+f(y)=f((x+y)/(1-xy))$ for all real x and y with $xy \neq 1$. —- Equation I

Solution:

Let $f(x)$ satisfy Equation I above. Differentiating Equation I partially with respect to each of x and y, we obtain $f^{'}(x)=((1+y^{2})/(1-xy)^{2})f^{'}((x+y)/(1-xy))$ Equation II $f^{'}(y)=((1+x^{2})/(1-xy)^{2})f^{'}((x+y)/(1-xy))$ Equation III

Eliminating common  terms in Equation II and Equation III, we deduce that $(1+x^{2})f^{'}(x)=(1+y^{2})f^{'}(y)$

As the left side of Equation III depends only on x and the  right side only on y, each side of Equation III must be equal to a constant c. Thus, we have $f^{'}(x)=\frac {c}{1+x^{2}}$

and so, $f(x)=c\arctan x +d$,

for some constant d. However, taking $y=0$ in Equation III, we obtain $f(x)+f(0)=f(x)$, so that $f(0)=0$ and $d=0$. Clearly, $f(x)=c\arctan x$ satisfies Equation I and so all solutions of Equation I are given by $f(x)=c\arctan x$ where c is a constant.

Note: In Equation II, the $f^{'}$ is w.r.t. x and in Equation III, the $f^{'}$ is w.r.t. y, yet in equation II $f^{'}(x+y)/(1-xy)$ is the same as the corresponding $f^{'}$ in Equation III because the argument $(x+y)/(1-xy)$ is symmetric w.r.t. x and y.