So I have this vector field $$F=\langle 2xe^{x^2y}, e^{x^2}, 1\rangle$$
I am asked to find a function such that $grad f=F$
However, when I check its curl, I got $$\begin{bmatrix} \vec i & \vec j & \vec k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ 2xe^{x^2y} & e^{x^2} & 1\\ \end{bmatrix}$$
I end up with $(2xe^{x^2}-2x^{3}e^{x^2y})\vec k \ne 0$
Since this vector is not conservative, I conclude that there's no such function $f$ where $grad f=F$.
But the question sounds very firm, did I miss something?