Find all functions $f(x+y)=f(x^{2}+y^{2})$ for positive $x,y$ Find all functions $f:\mathbb{R}^{+}\to \mathbb{R}$ such that for any $x,y\in \mathbb{R}^{+}$ the following holds:
$$f(x+y)=f(x^{2}+y^{2}).$$

If 

$$f(A(x,y))=f(B(x,y))$$ 

on a connected open set of $(x,y)$ where $A$ and $B$ are functionally independent (have nonzero Jacobian determinant), the function $f$ is constant on that set .
Locally one has both $xy$-coordinates and $AB$-coordinates.  From $(A,B)$ you can get to all close enough $(A + \epsilon_1, B + \epsilon_2)$ by moving $A$-only and then $B$-only along the $AB$ coordinates and this motion does not change the value of $f$. The motion can be tracked in $xy$ coordinates (uniquely determined by $A,B$ in a small neighborhood), and such paths can reach an open set of $(x,y)$ near any given point.    This shows $f$ is locally constant.  On a connected set that means constant.
The translation to this problem is: $A(x,y) = x+y$, $B(x,y)=x^2+y^2$, Jacobian is $\pm(x-y)$, so that the connected open set of $(x,y)$ with $0 < x < y$ can be used.  This is a subset of the allowed pairs but it covers all possible values of $A$ and $B$, and therefore of $f(A)$ and $f(B)$.  So it solves the problem with slightly less than the full set of assumptions, which is still a solution.
The need for an independence hypothesis can be seen from cases like $B=A^3$ where there are continuous nonconstant solutions on some intervals.
