<s>[INST] Finding all functions satisfying $f(x f(x+y))+f(f(y) f(x+y))=(x+y)^{2}$ Determine all functions $f:\mathbb{R} \to \mathbb{R}$ such that $$f(x f(x+y))+f(f(y) f(x+y))=(x+y)^{2}, \forall x,y \in \mathbb{R} \tag1)$$
My approach:
Let $x=0$, we get
$$f(0)+f\left((f(y))^2\right)=y^2$$
$\Rightarrow$
$$f\left((f(y))^2\right)=y^2-f(0)\tag2 $$
Let us assume $f(0)=k \ne 0$
Put $y=0$ above, we get
$$f(k^2)=-k$$
Also put $y=-x$ in $(1)$, we get
$$f(kf(x))+f(kf(-x))=0, \forall x \in \mathbb{R}$$
Put $x=0$ above we get
$$f(k^2)=0$$
$\Rightarrow$
$f(k^2)$ has two different images $0,-k$ which contradicts that $f$ is a function. Hence $k=0 \Rightarrow f(0)=0$.
So from $(2)$ we get:
$$f\left((f(y))^2\right)=y^2 \cdots (3)$$
Now put $y=0, x=f(x)$ in $(1)$, and use the fact $f(0)=0$,we get
$$f\left((f(x))^2\right)=(f(x))^2$$
Since $x$ is dummy variable, we get $$f\left((f(y))^2\right)=(f(y))^2 \cdots (4)$$
From $(3),(4)$, we get $$f(x)=\pm x$$
I just want to ask, is my approach fine? If not where is the flaw? Also other approaches are welcomed.
 [/INST] 
$f: \Bbb{R \to R}, f(xf(x+y))+f(f(y)f(x+y))=(x+y)^2.$

My attempt was to show that $f(0)=0$.. And I did it.
\begin{align}
P(0, y): \; & f(0)+f(f(y)^2)=y^2. \\
P(0, 0): \; & f(0)+f(f(0)^2)=0. \\
& \text{let } f(0)=k. \\
\ \\
\Rightarrow \; & k+f(f(y)^2)=y^2, k+f(k^2)=0. \\
\ \\
& \text{let } f(a)^2=f(b)^2. \\
P(0, a): \; & k+f(f(a)^2)=a^2. \\
P(0, b): \; & k+f(f(b)^2)=b^2. \\ 
\therefore \; & f(a)^2=f(b)^2 \Leftrightarrow a^2=b^2. \\
\ \\
P(k, -k): \; & f(kf(0))+f(f(-k)f(0))=0. \\
\Rightarrow \; & f(k^2)+f(kf(-k))=0, f(kf(-k))=k. \\
\ \\
P(-k, 0): \; & f(-kf(-k))+f(kf(-k))=k^2. \\
\Rightarrow \; & f(-kf(-k))=k^2-k. \\
P(k, -2k): \; & f(kf(-k))+f(f(-2k)f(-k))=k^2. \\
\Rightarrow \; & f(f(-2k)f(-k))=k^2-k. \\
\ \\
\Rightarrow \; & f(-kf(-k))=f(f(-2k)f(-k)). \\
\Rightarrow \; & f(-kf(-k))^2=f(f(-2k)f(-k))^2. \\
\ \\
\therefore \; & k^2f(-k)^2=f(-2k)^2f(-k)^2, k^2=f(-2k)^2. \\
\Rightarrow \; & f(0)^2=f(-2k)^2, 4k^2=0, k=0. \\
\therefore \; & f(0)=0. \\
\ \\
\ \\
\end{align}
</s>