$$ f(x)=x\cdot e^x \space ;\space x<-1 \\ W_{-1}(x)=f^{-1}(x) \\ W_{-1}(x\cdot e^x)=f^{-1}(x\cdot e^x)=x $$ But $W_{-1}(-\frac{2}{3}\cdot e^{-\frac{2}{3}}) = -1.429$ not $-0.66 \dot 6$?
It may be a stupid question, sorry about that.But please help I'm so confused.
$f:]-\infty, 0]\to [-1/e,0], x\mapsto x e^x$ is not injective. In fact, there are two different "branches" of the lambert W function. One is $W_0:[-1/e,0]\to[-1,0]$ and one is $W_{-1}:[-1/e,0]\to]-\infty,-1]$. They both satisfy $f\circ W_0=f\circ W_1=\operatorname{id}_{[-1/e,0]}$. You may think of this as the fact that the equation $$x e^x = -\frac 23 e^{-\frac 23}$$ has two distinct real solutions, one of them being $$W_0(-2/3 e^{-2/3})=-2/3$$ and the other being $$W_{-1}(-2/3 e^{-2/3})\approx -1.429.$$
Here is a desmos plot:
