Find the limit given that $f(1)=1$, $f(x+y)=f(x)+f(y)+2xy$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4}$ Given the function $f: \mathbb{R}\to \mathbb{R}$ satisfies:
\begin{cases}
f(1)=1 \\
f(x+y)=f(x)+f(y)+2xy,\quad&\forall x,y \in \mathbb{R}\\
f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4},&\forall x\neq0
\tag{1}\label{eqn1}
\end{cases}
Then find the limit of:
\begin{align}
L=\lim_{x \to 0} \frac{\frac{1}{e^{2f(x)}}-\sqrt[3]{1+f(x)}}{\ln\bigl(1+f(x)\bigr)}
\end{align}
My attempts
I can easily guess that $f(x)=x^2$, which totally satisfies \eqref{eqn1}. What I did was that I set $g(x)=f(x)-x^2$, then I got:
\begin{cases}
g(x)+g(y)=g(x+y)\\
g\left(\frac{1}{x}\right)x^4=g(x)
\tag{2}\label{eqn2}
\end{cases}
From \eqref{eqn1}, substitute $x=y=0$ we have $f(0)=0$
Now $g(1)=g(0)=0$, then I substitute $y=1-x$ in \eqref{eqn2} and got that
\begin{align}
g\left(\frac{1}{x}\right)x^4+g\left(\frac{1}{1-x}\right)(1-x)^4=0
\end{align}
Because the values of either $\frac1x$ or $\frac1{1-x}$ must lie between $0$ and $1$, I suspect that we can prove $g \equiv 0$.
Even if I have $f$ then, I'm still clueless about how to calculate the limit.
Any idea or solution or suggestion on the tools?
Any help is appreciated!

First to prove $g(x) = 0$:
We have $g(0) = 0$, $g(1) = f(1) -1^2 = 0$, so $g(x+1) = g(x)$. Also $g(x) + g(-x) = 0$ so $g(-x) = -g(x)$. We have:
$g(x) = g(x+1) = (x+1)^4 g(\frac{1}{x+1}) = -(x+1)^4 g(-\frac{1}{x+1}) = -(x+1)^4 g(1- \frac{1}{x+1}) = -(x+1)^4 g(\frac{x}{x+1})  = -(x+1)^4 (\frac{x}{x+1})^4 g(\frac{x+1}{x}) = -x^4 g(\frac{x+1}{x}) =-x^4  g(1 + \frac{1}{x}) = -x^4 g( \frac{1}{x}) = -x^4 \frac{1}{x^4} g(x)   = -g(x)$
Therefore $g(x) = 0$, $f(x) = x^2$. As for the limit, we can substitute $x^2$ with $y$ and apply  L'Hôpital's rule .
