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!

Calculating the given limit is standard, when you prove that $ f ( x ) = x ^ 2 $ for all $ x \in \mathbb R $. Let's focus on the nontrivial part of the problem. Using $ g ( 1 ) = 0 $,
$$ g ( x + y ) = g ( x ) + g ( y ) \text , \tag 0 \label 0 $$
and
$$ g \left ( \frac 1 x \right ) = \frac { g ( x ) } { x ^ 4 } \text , \tag 1 \label 1 $$
you can see that for any $ x \in \mathbb R \setminus \{ - 1 , 0 \} $,
\begin{align*}
\frac { g \left ( x ^ 2 \right ) } { x ^ 4 ( x + 1 ) ^ 4 } + \frac { g ( x ) } { x ^ 4 ( x + 1 ) ^ 4 }
& = \frac { g \left ( x ^ 2 + x \right ) } { \left ( x ^ 2 + x \right ) ^ 4 } \\
& \stackrel { \eqref {1} } = g \left ( \frac 1 { x ^ 2 + x } \right ) \\
& \stackrel { \eqref {0} } = g \left ( \frac 1 x \right ) - g \left ( \frac 1 { x + 1 } \right ) \\
& \stackrel { \eqref {1} } = \frac { g ( x ) } { x ^ 4 } - \frac { g ( x + 1 ) } { ( x + 1 ) ^ 4 } \\
& \stackrel { \eqref {0} } = \frac { g ( x ) } { x ^ 4 } - \frac { g ( x ) + g ( 1 ) } { ( x + 1 ) ^ 4 } \\
& = \frac { \left ( 4 x ^ 3 + 6 x ^ 2 + 4 x + 1 \right ) g ( x ) } { x ^ 4 ( x + 1 ) ^ 4 } \text ,
\end{align*}
which implies
$$ g \left ( x ^ 2 \right ) = 2 x ^ 5 ( x + 1 ) ^ 4 \left ( x ^ 2 + 3 x + 1 \right ) g ( x ) \text . \tag 2 \label 2 $$
Using \eqref{0} and \eqref{2} you get
\begin{align*}
& & & 2 x ^ 5 ( x + 1 ) ^ 4 \left ( x ^ 2 + 3 x + 1 \right ) g ( x ) + 2 g ( x y ) + 2 y ^ 5 ( y + 1 ) ^ 4 \left ( y ^ 2 + 3 y + 1 \right ) g ( y ) \\
& & \stackrel { \eqref {2} } = & g \left ( x ^ 2 \right ) + 2 g ( x y ) + g \left ( y ^ 2 \right ) \\
& & \stackrel { \eqref {0} } = & g \left ( ( x + y ) ^ 2 \right ) \\
& & \stackrel { \eqref {2} } = & 2 ( x + y ) ^ 5 ( x + y + 1 ) ^ 4 \left ( ( x + y ) ^ 2 + 3 ( x + y ) + 1 \right ) g ( x + y ) \\
& & \stackrel { \eqref {0} } = & 2 ( x + y ) ^ 5 ( x + y + 1 ) ^ 4 \left ( ( x + y ) ^ 2 + 3 ( x + y ) + 1 \right ) \bigl ( g ( x ) + g ( y ) \bigr ) \text ,
\end{align*}
which for $ y = 1 $ shows that
$$ \Bigl ( ( x + 1 ) ^ 5 ( x + 2 ) ^ 4 \left ( x ^ 2 + 5 x + 5 \right ) - x ^ 5 ( x + 1 ) ^ 4 \left ( x ^ 2 + 3 x + 1 \right ) - 1 \Bigr ) g ( x ) = 0 \text . \tag 3 \label 3 $$
Therefore, $ g ( x ) = 0 $ for all $ x \in \mathbb R \setminus A $, where $ A $ is a finite set of real numbers consisting of $ - 1 $, $ 0 $, and finitely many roots of the polynomial appearing on the left-hand side of \eqref{3}. For any $ a \in A $, choosing a positive integer $ n $ large enough so that $ a + n > b $ for all $ b \in A $, you can use \eqref{0} to see that $ g ( a ) = 0 $, and hence $ g ( x ) = 0 $ for all $ x \in \mathbb R $, or equivalently $ f ( x ) = x ^ 2 $ for all $ x \in \mathbb R $.
