Asymptote of non-algebraic implicit function

Question

The function f is implicitly given by $g(x,y)=e^{xy}-x-y=0$. I want to know about the asymptotes towards $+\infty$. I could deduce that there is only one intersect with the x-axis at (1|0) and by the implicit function theorem I know that apart from this point it consists of two continuously differentiable functions ($\partial_y g(x,y)\neq 0$ for $x>1$ and $f\in C^{\infty}$). So I decided to look at the two possibilities of $y>0$ and $y<0$(no crossing of the x-axis afterwards and continuous) for possible asymptotes and found that for $y(x)\rightarrow\frac{\ln(x)}{x}>0$ and $y(x)\rightarrow -x<0$ the condition $e^{xy}-x-y=0$ is fulfilled as the limit is concerned. Is this enough to say that those two functions have to be my asymptotes?

Answer 1

This is not an answer but probably too long for a comment.

The solution of $$e^{xy}-x-y=0$$ is explicitly given in terms of Lambert function $$y=-x-\frac{W\left(-x\,e^{-x^2} \right)}{x}$$ where $W(z)$ is Lambert function.

When $x$ start to be large, the argument becomes quite small and, around $z=0$ $$W(z)=z-z^2+O\left(z^3\right)$$ which makes $$y\sim -x+e^{-x^2}+x e^{-2 x^2} $$