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!

The easiest way to go is Taylor expansion.

Then the limit becomes
$$L=\lim_{x\to 0} \frac{e^{-2x^2}-\sqrt[3]{1+x²}}{\ln(1+x^2) } $$
$$=\lim_{x\to 0} \frac{(1-2x^2+2x^4-\cdots +\cdots )-(1+\frac{x^2}{3}-\frac{x^4}{9}+\cdots -\cdots)}{x^2-\frac{x^4}{2}+\frac{x^6}{3}+\cdots} $$
Simplifying this gives
$$L=\lim_{x\to 0} \frac{-\frac{7x^2}{3}+\frac{19x^4}{9}-\cdots +\cdots }{x^2-\frac{x^4}{2}+\frac{x^6}{3}+\cdots} $$
Dividing both numerator and denominator by $x^2$ gives the result
$$L=-\frac{7}{3}$$
