Normal functional equation: $ f \big( x f ( y ) + f ( x ) \big) + f \left( y ^ 2 \right) = f ( x ) + y f ( x + y ) $ 
Find all functions $ f : \mathbb R \to \mathbb R $ such that for any real numbers $ x $ and $ y $ we have
$$ f \big( x f ( y ) + f ( x ) \big) + f \left( y ^ 2 \right) = f ( x ) + y f ( x + y ) $$

What I tried:
$$ P ( 0 , - x ) \implies f \big( f ( 0 ) \big) + f \left( x ^ 2 \right) = f ( 0 ) - x f ( - x ) = f \left( x ^ 2 \right) $$
which implies that $ f $ is odd, because
$$ P ( 0 , x ) \implies f \big( f ( 0 ) \big) + f \left( x ^ 2 \right) = f ( 0 ) + x f ( x ) = f \left( x ^ 2 \right) $$
Also, considering
$$ P ( 1 , 1 ) \implies f \big( 2 f ( 1 ) \big) = f ( 2 ) \implies f ( 1 ) = 1 $$
Let $ a , b \in \mathbb R $ such that $ a ^ 2 = b $.
$$ P \left( a ^ 2 , a \right) \implies f \Big( a ^ 2 f ( a ) + f \left( a ^ 2 \right) \Big) = a f \left( a ^ 2 + a \right) \text . $$
Now with the observation that $ f ( - x ) = - f ( x ) $, we can deduce that $ f ( x ) = x $.
How to proceed?

Note that for the rest of this proof $a$ and $b$ will be any number in $\mathbb{R}$.
First, consider the following inputs. 
\begin{alignat}{3}
P(0,0) &\rightarrow f(f(0)) + f(0) = f(0) + 0 &&\rightarrow f(f(0)) = 0\\
P(0,1) &\rightarrow f(f(0)) + f(1) = f(0) + f(1) &&\rightarrow f(0) = f(f(0)) = 0\\
P(0,a) &\rightarrow f(f(0)) + f(a^2) = f(0) + af(a) &&\rightarrow f(a^2) = af(a)\\
P(0,-a) &\rightarrow f(f(0)) + f(a^2) = f(0) -af(-a) &&\rightarrow af(a) = -af(-a) &\rightarrow f(-a) = -f(a)\\
P(a,0) &\rightarrow f(af(0)+f(a)) + f(0) = f(a) + 0f(a) &&\rightarrow f(f(a)) = f(a)
\end{alignat}
These have yielded some equalities which we will use in the rest of this proof. Next off, we will prove that f(2a) = 2f(a). Consider the inputs (a,a) and (-a,a). These yield the following:
$$P(a,a) \rightarrow f((a+1)f(a)) + af(a) = f(a) + af(2a)$$
$$P(-a,a) \rightarrow f(-(a+1)f(a)) + af(a) = -f(a) + 2f(0) \rightarrow f((a+1)f(a)) = f(a) + af(a)$$
Note that these have the same factor $f((a+1)f(a))$. Setting the other sides equal yields:
$$f(a)+af(2a)-af(a) = f(a) + af(a) \rightarrow af(2a) = 2af(a) \rightarrow f(2a) = 2f(a)$$
Now, consider the inputs (a,b) and (-a,b). These yield the following:
$$P(a,b) \rightarrow f(af(b)+f(a)) + f(b^2) = f(a) + bf(a+b)$$
$$P(-a,b) \rightarrow f(-af(b)-f(a)) + f(b^2) = -f(a) + bf(b-a) \rightarrow \\ f(af(b)+f(a)) = f(a) +bf(b) +bf(a-b)$$
Note that both of these have the same term $f(af(b)+f(a))$. Setting these equal yields:
$$f(a) + bf(a+b) - bf(b) = f(a) + bf(b) + bf(a-b) \rightarrow bf(a+b) = 2bf(b) + bf(a-b) \rightarrow\\ f(a+b) = f(2b) + f(a-b)$$
Taking c = a-b and d=2b, this turns into:
$$f(c+d) = f(d) + f(c)$$
Using this equation on the starting equation yields:
\begin{align*}
&f(xf(y)+f(x)) + yf(y) = f(x) + yf(x+y) \rightarrow\\
&f(xf(y)) + f(f(x)) + yf(y) = f(x) + yf(x) + yf(y) \rightarrow\\
&f(xf(y)) = yf(x)
\end{align*}
Putting $P(1,a)$ into this new equation yields $f(f(a)) = af(1) \rightarrow f(a) = af(1)$. In other words, $f$ is linear. That means that the only functions which could still be $f$, are of the form $f(x) = ax$. Note, however, that we also have $$f(f(x)) = f(x) \rightarrow a(a(x)) = a(x) \rightarrow a^2x = ax \rightarrow a = 1 \lor a = 0$$
In other words, the only functions which statisfy this equation, are $f(x) = 0$ and $f(x) = x$.
