<s>[INST] Solving $f(yf(x)+x/y)=xyf(x^2+y^2)$ over the reals 
Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that $f(1)=1$ and for all real numbers $x$ and $y$ with $y \neq 0$,  $$f\Bigg (yf(x)+\frac{x}{y}\Bigg)=xyf(x^2+y^2)$$

This seems quite hard. $f(x)=\begin{cases}\frac{1}{x}, x\neq 0 \\ 0, x=0 \end{cases}$ works by inspection, as does $f(x)=0$ (though I'm not sure if this is legitimate. If we set $y=1$, we have $f(f(x)+x)=xf(x^2+1)$. If we set $x=1$, we have $f(\frac{y^2+1}{y})=yf(y^2+1)$. which seems to imply $f(f(x)+x)=f(\frac{x^2+1}{x})$. If I could show that $f$ is injective I could go further, but I'm stuck - subbing in other values doesn't really seem to lead anywhere either. 
I believe this problem came from an Olympiad camp. 
 [/INST] Let $f:\textbf{R}\rightarrow\textbf{R}$, such that $f(1)=1$ and
$$
f\left(yf(x)+\frac{x}{y}\right)=xyf(x^2+y^2)\textrm{, }\forall (x,y)\in\textbf{R}\times\textbf{R}^{*}\tag 1
$$
Set $x=0$, $y=1$ in (1), then easily $f(0)=0$ (the case $f(0)\neq 0$ is trivial). For $y=1$ in (1) we get
$$
xf(x^2+1)=f(f(x)+x)\textrm{, }x\neq 0.\tag 2
$$ 
Also using the symmetry we get
$$
f\left(yf(x)+\frac{x}{y}\right)=f\left(xf(y)+\frac{y}{x}\right)\textrm{, }x,y\neq 0
$$
For $y=1$, we get
$$
f\left(x+\frac{1}{x}\right)=f(f(x)+x)\textrm{, }x\neq 0\tag 3
$$
Assume that $h(x,y)$ is a surface (function) such that 
$$
f(y+h(x,y))=f(x).\tag 4
$$
Note. 
One choise of $h(x,y)$ is $h(x,y)=-y+x$ but may be there others. For example if $f(x)=x^2-x+1$, then $h(x,y)=x-y$ or $h(x,y)=1-x-y$.
We assume here that 
$$
f(x)=h\left(x+\frac{1}{x},x\right)\tag{4.1}
$$
From (4) with $x\rightarrow x+\frac{1}{x}$ we get 
$$
f\left(x+\frac{1}{x}\right)=f\left(y+h\left(x+\frac{1}{x},y\right)\right).
$$
Hence for $y=x$ in the above identity we get 
$$
f\left(x+\frac{1}{x}\right)=f\left(x+h\left(x+\frac{1}{x},x\right)\right)=f(x+f(x))
$$
Hence we can get (3). From (2) and (3) we get also
$$
xf(x^2+1)=f\left(x+\frac{1}{x}\right)\tag 6
$$
Setting $x\rightarrow x^{-1}$ in (6)
$$
\frac{1}{x}f\left(1+\frac{1}{x^2}\right)=f\left(x+\frac{1}{x}\right)=xf(x^2+1).\tag 7
$$
Hence if we set $x^2\rightarrow x>0$ in (7), then
$$
f\left(1+\frac{1}{x}\right)=xf(x+1)\textrm{, for all }x>0.\tag 8
$$
Set also $x=y-1>0$ in (8), then 
$$
f\left(\frac{y}{y-1}\right)=(y-1)f(y)\textrm{, }y>1.
$$
With $y=1/w>1$ we get
$$
\frac{1}{1-w}f\left(\frac{1}{1-w}\right)=\frac{1}{w}f\left(\frac{1}{w}\right).
$$
Hence if $0<w<1$ and $g(w):=\frac{1}{w}f\left(\frac{1}{w}\right)$, then
$$
f(x)=x^{-1}g\left(x^{-1}\right)\textrm{, where }x>1\textrm{ and }g(1-w)=g(w)\textrm{, }0<w<1\tag 9
$$
The solution (9) satisfies (8),(7),(6), but for to holds (3) we get plus a new functional equation for $f(x)$ and this is (4.1).
Hence the general solution is
$$
f(x)=h\left(x+\frac{1}{x},x\right)\tag{11}
$$
with
$$
\frac{1}{x-1}h\left(x-1+\frac{1}{x-1},\frac{1}{x-1}\right)=\frac{1}{x}h\left(x+\frac{1}{x},\frac{1}{x}\right)\tag{12}
$$
and $h(x,y)$ solution of 
$$
f(y+h(x,y))=f(x).\tag{13}
$$ 
An obvious solution of $(13)$ is $h(x,y)=x-y$, but it may exist and other solutions. 
</s>