Basically, I just want to show that the right derivative of $W(-xe^{-x})$ at $x=1$ is $1$:
WolframAlpha: plot lambertW(-x*e^(-x)) for x=0 to 2
Equivalently, I want to prove the following limit:
$$ \lim_{x\to 1^+} \frac{W\left(-xe^{-x}\right)+1}{x-1} = 1 $$
WolframAlpha: limit x->1+ of (lambertW(-x*exp(-x))+1)/(x-1)
L'Hôpital doesn't help, because it goes back to the same limit.
It's clearer if you plot the second real branch too (I set $x = -x$):
If $y = W_k(x e^x)$, then $F(x, y) = y e^y - x e^x = 0$. The second-order terms in the Taylor expansion of $F$ around $(-1, -1)$ are $C (y + 1)^2 - C (x + 1)^2$, therefore the tangents to the curve $F = 0$ are $y = x$ and $y = -x -2$.