Arc length integral for the curve $e^{2x+2y}=x-y$. Arc length integral for the curve 

$$e^{2x+2y}=x-y$$

So I tried isolating $y$ or $x$ and got stuck with a $0 = 0$ at the end of it.
I tried to also use partial derivatives. Am I going in the right direction?
 A: Hint: Rotate your axes by $\pi/4$
A: $\displaystyle 2(x-y)e^{2(x-y)}=2e^{4x}\enspace$ => $\displaystyle \enspace x-y=\frac{1}{2}W(2e^{4x})$ 
where $W(x)$ is the Lambert W-function
We get $\enspace\displaystyle \frac{dy}{dx}=\frac{1-W(2e^{4x})}{1+W(2e^{4x})}$ .
For the arc length integral follows: 
$\displaystyle \int\sqrt{(dx)^2+(dy)^2}=\int\sqrt{1+(\frac{1-W(2e^{4x})}{1+W(2e^{4x})})^2}dx=\sqrt{2}\int\frac{\sqrt{1+W(2e^{4x})^2}}{1+W(2e^{4x})}dx$
$\displaystyle =\frac{\sqrt{2}}{4}( \sqrt{1+W(2e^{4x})^2}-\ln(1+\sqrt{1+W(2e^{4x})^2})+\ln(W(2e^{4x})))+C$
$\displaystyle =\frac{\sqrt{2}}{4}( \sqrt{1+(2x-2y)^2}-\ln(1+\sqrt{1+(2x-2y)^2})+\ln(2x-2y))+C$
A: For $e^{2x+2y}= x-y$ let variable changing $u=2x+2y$ and $v=x-y$ the curve is $e^u=v$ so Jacobin is $J=-2$ and the arc length integral for this curve from $(a,A)$ to $(b,B)$ is
\begin{eqnarray}
\ell 
 &=&
\int_a^b\sqrt{1+y'^2}dx\\
 &=&
\int_{\alpha}^{\beta}\sqrt{1+v'^2}\,|J|\,du\\
 &=&
\int_{\alpha}^{\beta}2\sqrt{1+e^{2u}}du\\
 &=&
\left(2\sqrt{1+e^{2u}}-2arctanh\sqrt{1+e^{2u}}\right)\Big|_{\alpha}^{\beta}\\
 &=&
\left(2\sqrt{1+e^{2x+2y}}-2arctanh\sqrt{1+e^{2x+2y}}\right)\Big|_a^b
\end{eqnarray}
