We have $\vec{F}(x,y,z) = (2xcos(y)-2z^3)i + (3+2ye^z-x^2sin(y)) j + (y^2e^z-6xz^2) k$
Now a potential field $\vec{F}(x,y,z) = grad(\phi)(x,y,z) = f_x(x,y,z)i+f_y(x,y,z)j+f_z(x,y,z)k$
But after taking the partials $\vec{F}(x,y,z) \neq grad(\phi)(x,y,z)$
Since $f_x(x,y,z)i$ = $2cos(y)-2z^3$ $f_y(x,y,z)j = 3+2e^z-x^2 sin(y)$ and $f_x(x,y,z)k = y^2e^z-6x2z$ which is not equal to our original $\vec{F}(x,y,z)$
So $\phi(x,y,z) = 2cos(y)-2z^3 + 3+2e^z-x^2 sin(y) + y^2e^z-6x2z \neq \vec{F}(x,y,z)$ $ = (2xcos(y)-2z^3)i + (3+2ye^z-x^2sin(y))j + (y^2e^z-6xz^2)k$
$\vec{F}(x,y,z) = (2x \cos y - 2z^3) \hat i + (3+2ye^z-x^2 \sin y) \hat j + (y^2e^z-6xz^2) \hat k$
Checking, the curl of the vector field is $(0, 0, 0)$ so it is a conservative vector field. To find potential function, we first integrate $\hat i$ component of the vector field with respect to $dx$.
$f(x, y, z) = x^2 \cos y - 2 x z^3 + \int g(y, z) \ dy$
Taking $\hat j$ component, $g(y, z) = 3 + 2 y e^z$
Now integrating $g(y, z)$ with respect to $dy$,
$f(x, y, z) = x^2 \cos y - 2 x z^3 + 3 y + y^2 e^z + \int h(z) \ dz$
Checking $\hat k$ component, $h(z) = 0$
So $f(x, y, z) = x^2 \cos y - 2 x z^3 + 3 y + y^2 e^z$