Solving $(x+y) \exp(x+y) = x \exp(x)$ for $y$. While thinking about the Lambert $W$ function I had to consider
Solving $(x+y) \exp(x+y) = x \exp(x)$ for $y$.
This is what I arrived at:
(for $x$ and $y$ not zero)
$(x+y) \exp(x+y) = x \exp(x)$
$x\exp(x+y) + y \exp(x+y) = x \exp(x)$
$\exp(y) + y/x \exp(y) = 1$
$y/x \exp(y) = 1 - \exp(y)$
$y/x = (1-\exp(y))/\exp(y)$
$x/y = \exp(y)/(1-\exp(y))$
$x = y\exp(y)/(1-\exp(y))$
$1/x = 1/y\exp(y) -1/y$
And then I got stuck.
Can we solve for $y$ by using Lambert $W$ function? 
Or how about an expression with integrals?

The solution of $ (x+y) \exp(x+y) = x \exp(x) $ is given in terms of the Lambert W function
Let $z=x+y$, then we have
$$ z {\rm e}^{z} = x {\rm e}^{x} \Rightarrow z = { W} \left( x{{\rm e}^{x}} \right) \Rightarrow y = -x + { W} \left( x{{\rm e}^{x}} \right)  \,. $$ 
Added: Based on the comment by Robert, here are the graphs of $ y = -x + { W_0} \left( x{{\rm e}^{x}} \right)  $ and $ y = -x + { W_{-1}} \left( x{{\rm e}^{x}} \right)  $


