Answer
Conservative; $f(x,y,z)=e^{x} \sin yz+K$
Work Step by Step
The vector field $F$ will be conservative if and only if $curl F=0$
consider $F=A i+B j+C k$
$curl F=[C_y-B_z]i+[A_z-C_z]j+[B_x-A_y]k$
Here, we have $curl F=[(e^{x}\cos yz-yze^{x}\sin yz)-(e^{x} \cos yz-yze^{x} \sin yz)]i+[(ye^{x} \cos yz-ye^{x} \cos yz)]j+[(ze^{x} \cos yz-ze^{x} \cos yz)-k=0$
Thus, the vector field $F$ is conservative.
$f(x,y,z)=e^{x} \sin yz+g(y,z)$
$f_y=z e^{x} \cos yz+g_y \implies g'(y)=0$
Thus, $g_y=h(z)$ and $f_y=z e^{x} \cos yz$
Now, $f(x,y,z)=e^{x} \sin yz+h(z)$
Thus, $f_z=y e^{x} \cos yz+h'(z)$
$ \implies h'(z)=0$
Hence, $f(x,y,z)=e^{x} \sin yz+K$
