A functional equation problem: $ \frac { f \left( f ( y ) ^ 2 + x y \right) } { f ( y ) } = f ( x ) + y $ 
Let $ \mathbb R ^ + $ denote the set of the positive real numbers. Find all functions $ f : \mathbb R ^ + \to \mathbb R ^ + $ satisfying
$$ \frac { f \left( f ( y ) ^ 2 + x y \right) } { f ( y ) } = f ( x ) + y $$
for all $ x , y \in \mathbb R ^ + $.

I am very thankful for any solution, please help!
I tried to set $ x = y = 1 $, $ x = y = 2 $, $ x = 1 $, $ y = 2 $, so on, but this problem is more difficult.
 A: $ \def \R {\mathbb R ^ +} $
You can show that the only $ f : \R \to \R $ satisfying
$$ f \left( x y + f ( y ) ^ 2 \right) = \big( f ( x ) + y \big) f ( y ) \tag 0 \label 0 $$
for all $ x , y \in \R $ is the identity function. It's straightforward to verify that the identity function is a solution. We try to prove the converse. The messiest part is showing $ f ( 1 ) = 1 $, and after that, everything goes rather smoothly. Before going further, let's make the simple observation that $ f $ must be injective: if $ f ( x ) = f ( y ) $, then by \eqref{0} we have
$$ x = \frac { f \left( y x + f ( x ) ^ 2 \right) } { f ( x ) } - f ( y ) = \frac { f \left( x y + f ( y ) ^ 2 \right) } { f ( y ) } - f ( x ) = y \text . $$

Showing $ f ( 1 ) = 1 $
Letting $ a = f ( 1 ) $ and plugging $ y = 1 $ in \eqref{0} we get
$$ f \left( x + a ^ 2 \right) = a \big( f ( x ) + 1 \big) \text . \tag 1 \label 1 $$
Assuming $ a > 1 $, \eqref{1} gives
$$ f \left( x + n a ^ 2 \right) = a ^ n f ( x ) + \frac { a ^ { n + 1 } - a } { a - 1 } \tag 2 \label 2 $$
by induction on the positive integer $ n $. Setting $ x = y = 1 + n a ^ 2 $ in \eqref{0} and using \eqref{2} we get
$$ f \left( \left( 1 + n a ^ 2 \right) ^ 2 + \left( \frac { a ^ { n + 2 } - a } { a - 1 } \right) ^ 2 \right) = \frac { a ^ { n + 2 } - a } { a - 1 } \left( \frac { a ^ { n + 2 } - a } { a - 1 } + 1 + n a ^ 2 \right) \text . \tag 3 \label 3 $$
Also, letting $ m = 2 n + \left\lfloor n ^ 2 a ^ 2 + \left( \frac { a ^ { n + 1 } - 1 } { a - 1 } \right) ^ 2 \right\rfloor $ and $ b = \left( 1 + n a ^ 2 \right) ^ 2 + \left( \frac { a ^ { n + 2 } - a } { a - 1 } \right) ^ 2 - m a ^ 2 $, we can see that $ m $ is a positive integer and $ b \in \R $, and thus we can use \eqref{2} to get
$$ f \left( \left( 1 + n a ^ 2 \right) ^ 2 + \left( \frac { a ^ { n + 2 } - a } { a - 1 } \right) ^ 2 \right) = a ^ m f ( b ) + \frac { a ^ { m + 1 } - a } { a - 1 } \text . $$
But this means that the right-hand side of \eqref{3} must be greater than $ \frac { a ^ { m + 1 } - a } { a - 1 } $, which can't happen for large enough $ n $. Thus the assumption $ a > 1 $ is absurd, and we must have $ a \le 1 $. Now, if $ a < 1 $, we can set $ x = 1 - a ^ 2 $ and $ y = 1 $ in \eqref{0} to get $ a = a \Big( f \left( 1 - a ^ 2 \right) + 1 \Big) $, which contradicts $ a , f \left( 1 - a ^ 2 \right) \in \R $. Thus the assumption $ a < 1 $ is absurd, too, and we must have $ a = 1 $.

Showing $ f = \operatorname {Id} _ { \R } $
Using \eqref{1} and induction on the positive integer $ n $, we get
$$ f ( x + n ) = f ( x ) + n \text , \tag 4 \label 4 $$
which in particular shows $ f ( n ) = n $ for any positive integer $ n $. Hence, letting $ y = n $ in \eqref{0} we have
$$ f ( n x ) = n f ( x ) \tag 5 \label 5 $$
for any positive integer $ n $. Letting $ x = \frac n y $ in \eqref{0} and using \eqref{4} and \eqref{5}, we get
$$ f \left( f ( y ) ^ 2 \right) - y f ( y ) = n \Bigg( f ( y ) f \left( \frac 1 y \right) - 1 \Bigg) \text , $$
and since the left-hand side doesn't depend on $ n $, both sides must be equal to $ 0 $. In particular, we have
$$ f \left( f ( y ) ^ 2 \right) = y f ( y ) \text . \tag 6 \label 6 $$
Letting $ x = \frac { f ( y ) ^ 2 } y $ in \eqref{0} and using \eqref{5} and \eqref{6} we get
$$ f \left( \frac { f ( y ) ^ 2 } y \right) = y \text . \tag 7 \label 7 $$
Using \eqref{4} and \eqref{7}, we can see that
$$ f \left( \frac { f ( y ) ^ 2 + 2 f ( y ) + 1 } { y + 1 } \right) = f \left( \frac { \big( f ( y ) + 1 \big) ^ 2 } { y + 1 } \right) = f \left( \frac { f ( y + 1 ) ^ 2 } { y + 1 } \right) \\
= y + 1 = f \left( \frac { f ( y ) ^ 2 } y \right) + 1 = f \left( \frac { f ( y ) ^ 2 } y + 1 \right) = f \left( \frac { f ( y ) ^ 2 + y } y \right) \text . $$
By injectivity of $ f $, we have
$$ \frac { f ( y ) ^ 2 + 2 f ( y ) + 1 } { y + 1 } = \frac { f ( y ) ^ 2 + y } y \text , $$
or equivalently
$$ y f ( y ) ^ 2 + 2 y f ( y ) + y = y f ( y ) ^ 2 + y ^ 2 + f ( y ) ^ 2 + y \text , $$
which can be simplified to $ \big( f ( y ) - y \big) ^ 2 = 0 $, and thus $ f $ is the identity function.
A: Put $y=0$. Then the equation shows that $f$ has to be constant. Say $f(x)=c$ for all $x$. Using this in the equation results in $1=c+y$ for all $y>0$ which is impossible. Thus the equation has no solution.
EDIT: As pointed out below the function is defined on $\mathbb{R}_+$. Thus the arguments are not valid. 
A: Proving with  $f(x)=kx^n$ we have$$f(k^2y^{2n}+xy)=k(k^2y^{2n}+xy))^n\\f(y)(f(x)+y)=k(f(y)f(x)+yf(y))=k(k^2y^nx^n+ky^{n+1})$$ which is valid only for $k=n=1$
We have an example with $$f(x)=x$$ I hope to find other examples or prove that there are not
