Proving Abel's identity for the Dilogarithm. Abel's Identity for the dilogarithm is stated as follows

Theorem (Abels Identity)
Given that $x,y \,{\not\in}\,[1,\infty)$ then
  \begin{align*}
 \log(1-x) \log(1-y) 
                = 
\operatorname{Li}_2(u) + \operatorname{Li}_2(v) - \operatorname{Li}_2(u v) - \operatorname{Li}_2(x) - \operatorname{Li}_2(y)\,
 \end{align*}
  holds for all $x,y$. Where $u = \frac x{1-y}$ and $v = \frac y{1-x}$.

I was able to find the original paper where the statement is proven. It can be read here Abels Identity. I was very pleased finding the paper, and being from Norway I thought I should be able to read it. (No pun inteded.) Alas the paper is written in french, a language foreign to me. Can anyone help me outline the proof for this identity?
I tried differentiating the expression with regard to both $x$ and $y$ but it all got very messy, very fast. All help is appreciated =)
 A: I don't understand why do you have troubles when differentiating. Using that
$$1-u=\frac{1-x-y}{1-y},\qquad 1-v=\frac{1-x-y}{1-x},\qquad 1-uv=\frac{1-x-y}{(1-x)(1-y)}$$
and that $\displaystyle\mathrm{Li}_2'(x)=-\frac{\ln(1-x)}{x}$, the derivative of the right hand side with respect to $x$ is
\begin{align}
&-\ln(1-u)\left(\ln u\right)'_x-\ln(1-v)\left(\ln v\right)'_x+\ln(1-uv)\left(\ln u+\ln v\right)'_x+\frac{\ln(1-x)}{x}=\\
&=-\frac{\ln(1-x-y)-\ln(1-y)}{x}-\frac{\ln(1-x-y)-\ln(1-x)}{1-x}+\\
&+\Bigl(\ln(1-x-y)-\ln(1-x)-\ln(1-y)\Bigr)\left(\frac{1}{x}+\frac{1}{1-x}\right)+\frac{\ln(1-x)}{x}.
\end{align}
In the last expression, the terms containing $\ln(1-x-y)$ and $\ln(1-x)$ obviously cancel out so that it reduces to
$$-\frac{\ln(1-y)}{1-x}=\frac{d}{dx}\ln(1-x)\ln(1-y).$$
Therefore, to show the identity, it now suffices to verify it for some value of $x$. For example, for $x=0$ we have $u=0$, $v=y$ and the right hand side becomes
$$\mathrm{Li}_2(0)+\mathrm{Li}_2(y)-\mathrm{Li}_2(0)-\mathrm{Li}_2(0)-\mathrm{Li}_2(y)
=-\mathrm{Li}_2(0)=0.$$
A: This is a direct translation of Abel's nice paper 'Note sur la fonction $\psi x=x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+\cdots$'.
Start with $\;\displaystyle\operatorname{Li}_2(x)=-\int\frac{\log(1-x)}x\,dx\;$ 
and set $\;\displaystyle x:=\frac a{1-a}\frac y{1-y}\;$ with '$a$' constant.
Then $\;\log(x)=\log(a)-\log(1-a)+\log(y)-\log(1-y)\,$ and the differentials will verify :
$$\frac {dx}x=\frac {dy}y+\frac {dy}{1-y}$$
Since $\,(1-a)(1-y)-ay=1-a-y\,$ we get :
(factorizing further $(1-y)$ in the logarithm for $\frac{dy}y\;$ and $(1-a)$ for $\frac{dy}{1-y}$)
\begin{align}
\operatorname{Li}_2\left(\frac a{1-a}\frac y{1-y}\right)&=-\int\left(\frac{dy}y+\frac{dy}{1-y}\right)\log\frac{1-a-y}{(1-a)(1-y)}\\
&=-\int\frac{dy}y\log\left(1-\frac y{1-a}\right)+\int \frac{dy}y\log(1-y)\\
&\quad-\int\frac{dy}{1-y}\log\left(1-\frac a{1-y}\right)+\int \frac{dy}{1-y}\log(1-a)\\
\end{align}
But the integrals at the right may be written as $\operatorname{Li}_2$ functions since :
\begin{align}
&\int\frac{dy}y\log\left(1-\frac y{1-a}\right)=-\operatorname{Li}_2\left(\frac y{1-a}\right),\\
&\int\frac{dy}y\log\left(1-y\right)=-\operatorname{Li}_2\left(y\right);\\
\end{align}
so that $$ \operatorname{Li}_2\left(\frac a{1-a}\frac y{1-y}\right)=\operatorname{Li}_2\left(\frac y{1-a}\right)-\operatorname{Li}_2\left(y\right)-\log(1-a)\log(1-y)-\int\frac{dy}{1-y}\log\left(1-\frac a{1-y}\right)$$
Set $z:=\dfrac a{1-y}$ i.e. $1-y=\dfrac az,\;dy=\dfrac {a\,dz}{z^2}$ to get :
$$\int\frac{dy}{1-y}\log\left(1-\frac a{1-y}\right)=\int\frac{dz}z\log(1-z)=-\operatorname{Li}_2\left(z\right)=-\operatorname{Li}_2\left(\frac a{1-y}\right)$$
$$\operatorname{Li}_2\left(\frac a{1-a}\frac y{1-y}\right)=\operatorname{Li}_2\left(\frac y{1-a}\right)+\operatorname{Li}_2\left(\frac a{1-y}\right)-\operatorname{Li}_2\left(y\right)-\log(1-a)\log(1-y)+C$$
The arbitrary constant $C$ will be determined by $\,y=0$ to get $\,C=-\operatorname{Li}_2(a)$.
Replacing $a$ by $x$ we get :
$$\operatorname{Li}_2\left(\frac x{1-x}\frac y{1-y}\right)=\operatorname{Li}_2\left(\frac y{1-x}\right)+\operatorname{Li}_2\left(\frac x{1-y}\right)-\operatorname{Li}_2\,y-\operatorname{Li}_2\,x-\log(1-x)\log(1-y)$$
Set $\;u:=\dfrac x{1-y},\;v:=\dfrac y{1-x}\,$ to conclude :
$$\operatorname{Li}_2(u v)=\operatorname{Li}_2(u) + \operatorname{Li}_2(v) - \operatorname{Li}_2(x) - \operatorname{Li}_2(y)-\log(1-x) \log(1-y) $$
A: For the proof of li_2(x) +li_2(-x)  =½li_2(x²)
rewrite the dilogs at the right hand side as sums
  We have €(x^n/n²) +((-x) ^n/n²) from 1 to infinity. 
Then notice the odd terms keep cancelling out. 
Leaving only the even terms so rewrite sum
   €(x^2n/4n² +(-x) ^2n/4n²)
   2/4€(x^2n/n²)
Which is recognized as 
½li_2 (x²)
