$$f(x) = \frac{e^{1-\ln(x-x^2)}}{\ln(1 - e^{x-x^2})}$$
I was solving this, and I found the domain is the emptyset.
Yet, then I checked with Mathematica and it returned me a different domain. When I tried to check the functions separately, it said that the domain of the numerator is $x < 0$ union $0 < x < 1$ union $x > 1$.
I think that this is beacuse it sees
$$e^{1-\ln(x-x^2)} = e\cdot e^{-\ln(x-x^2)} = \frac{e}{x-x^2}$$
But isn't this wrong?? The last property only holds when the argument of the logarithm is positive, isn't it?
Check the conditions for the function definition. If you consider the arguments of the logarithmic functions in the numerator and denominator:
Numerator:
$\ln(x - x^2)$ requires $x - x^2 > 0$
Solving $x - x^2 > 0$ we get $x(1 - x) > 0$.
The solutions for $x(1 - x) = 0$ are $x = 0$ and $x = 1$, therefore the inequality $x - x^2 > 0$ holds when $0 < x < 1$.
Denominator:
The exponential function $e^{x-x^2}$ is defined for all real $x$, meaning there are no restrictions coming from here.
However, $1 - e^{x-x^2} > 0$. So we solve for $e^{x-x^2} < 1$. The expoential function is equal to 1 when $x-x^2 = 0$ and less then 1 when $x-x^2 < 0$.
Conclusion:
There is no value or range of $x$ where $x-x^2 < 0$ and $0 < x < 1$, so the function $f(x)$ indeed has no domain where it is defined, as no value of $x$ can satisfy the numerator and denominator conditions simulteneously.