<s>[INST] If $y = e^{x(y-1)}$ then $y \approx 1-2(x-1)$ when $0Assume $y = e^{x(y-1)}$. 
Then $y \approx 1-2(x-1)$ when $0<x-1<<1$

I thought of something like that:
$$ e^{x(y-1)} = e^{-2(x-1)}e^{xy+x-2}=(1-2(x-1)+O(x-1)^2)e^{xy+x-2}$$
But I failed to prove that $e^{xy+x-2} \approx 1$
Any ideas?
Thanks
 [/INST] This isn't rigorous, but will give you a starting point: since $e^z \approx 1 + z + \dfrac 12 z^2$ you have $y = e^{x(y-1)} \approx 1 +x(y-1) + \dfrac 12 x^2 (y-1)^2$. That is, $$y-1 \approx x(y-1) + \frac 12 x^2 (y-1)^2$$ so that (after canceling the $y-1$) $$1 \approx x + \dfrac 12 x^2 (y-1)$$ and $$y \approx \frac{2}{x^2} (1-x) + 1$$ Since $0 < x-1 << 1$ you can approximate $\dfrac 2{x^2}$ by $2$.
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