Applying the implicit function theorem on $e^y+x^2ye^{-x}=3$

Question

I know the implicit function theorem but no clue how to show this..

Show that $e^y+x^2ye^{-x}=3$ implicitly defines a unique function $y=f(x)$ defined on all real numbers.

Answer 1

Hint: note that $x^2 e^{-x} \ge 0$, and so for each $x$, the function $y \mapsto e^y + x^2e^{-x}y$ is strictly increasing.