I came across a function: $$f(x,y)=\frac{x}{y^{3}}\left[\left(1-e^{-y x}\right)^{2}- 3\right]+\frac{3}{2y^{4}}\left[\left(2-e^{-y x}\right)^{2}- 1\right]$$
I am not sure if $f(x,y)<0$ in the domain $x>1$ and $y\in \mathbb{R}$. I have used Excel to obtain some numerical result and guess $f(x,y)<0$. Am I correct? Please help me for this question.
Your function $f$ is indeed negative for all $x\ne0$.
Proof. One has $$f(x,y)=-{x^4\over2} g(xy)$$ with $$g(z):={(12+4z)e^{-z}-(3+2z)e^{-2z}-(9-4z)\over z^4}={1\over2}-{7z\over15}+{z^2\over4}-{31 z^3\over 315}+\ldots\tag{1}$$ an entire function of the single variable $z$ (the singularity at $z=0$ is removable). Let $$h(z):=(12+4z)e^{-z}-(3+2z)e^{-2z}-(9-4z)$$ be the numerator in $(1)$. Then $$h(0)=h'(0)=0\tag{2}$$ and $$h''(z)=4e^{-2z}p(z),\qquad p(z):=(1+z)e^z-1-2z\ .$$ It follows that $h''(0)=4p(0)=0$. Furthermore $$p'(z)=ze^z+2(e^z-1)$$ is obviously negative for negative $z$ and positive for positive $z$. This implies that $p$ and hence $h''$ is positive for all $z\ne0$. Together with $(2)$ we can conclude that $h$ is positive for all $z\ne0$, and $(1)$ then shows that $g$ is positive for all real $z$.$\qquad\square$