Finding $f(x)$ from the functional equation $f(x+2y)=f(x)+f(2y)+e^{x+2y}(x+2y)-xe^x-2ye^{2y}+4xy$ This sum really looks scary to me. No idea how to begin. I just figured out that $f(0)$ is $0$. What's next?

Let $f:\mathbb R \to \mathbb R$, such that $f'(0)=1$ and
$$f(x+2y)=f(x)+f(2y)+e^{x+2y}(x+2y)-xe^x-2ye^{2y}+4xy$$
for all real $x$ and $y$. Find $f(x)$.

P.S. Please don't use very high level maths.
 A: Set $y = h/2$ and $g(x) = xe^x$, then
$$
\frac{f(x+h) - f(x)}h = \frac{f(h)}h + \frac{g(x+h)-g(x)}h
-e^{h}+2x
$$
Now take the limit for $h \to 0$ (and use $f(0)=0$ which you already obtained):
$$
\frac{f(x+h) - f(x)}h \to f'(x) \, , \quad \frac{f(h)}h = \frac{f(h) - f(0)}{h-0} \to f'(0) \, , \quad \text{etc.}
$$
It follows that
$$
 f'(x) = f'(0) + g'(x) - 1 + 2x = g'(x) +2x \, .
$$
Together with $f(0) = 0$ it follows that
$f(x) = g(x) + x^2 = xe^x + x^2$.
A: Partial differentiating the given equation w.r.t x keeping y constant we get
$$ f'(x+2y)=f'(x)+ e^{x+2y}(x+2y)+e^{x+2y}-e^x-e^xx+4y$$
Now put $x=0$ and $y=x/2$ to get
$$ f'(x)=e^xx+e^x+2x $$
Now integrate both sides and find the constant by putting $x=0$ to get $$ f(x)=xe^x+x^2 $$
A: OK, I prefer to replace $2y$ with $y$ to obtain
$$f(x+y)=f(x)+f(y)+(x+y)e^{x+y}-xe^x-ye^{y}+2xy \tag{1}$$
Next, choose $x=y$ to obtain
$$f(2x)=2f(x)+2xe^{2x}-2xe^{x}+2x^2 \tag{2}$$
Or equivalently
$$\begin{align}
& f(2x)-2f(x)=(2xe^{2x}+(2x)^2)-2(xe^x+x^2) \\
& f'(0)=1
\end{align}
\tag{3}$$
Now, I will prove that the only function satisfying $(3)$ is
$$f(x)=xe^x+x^2 \tag{4}$$
which is suggestive when looking at $(3)$.
Suppose that there is another function called $g(x)$ satisfying
$$\begin{align}
& g(2x)-2g(x)=(2xe^{2x}+(2x)^2)-2(xe^x+x^2) \\
& g'(0)=1
\end{align}
\tag{5}$$
Next, introducing $h(x)=f(x)-g(x)$ we conclude that
$$\begin{align}
& h(2x)=2h(x) \\
& h'(0)=0
\end{align}
\tag{6}$$
and so $h(x)=0$ leading to $f(x)=g(x)$. For a proof of $h(x)=0$ on $\mathbb R$ see this post.
