Answer
$f(x,y,z)=e^{x} \sin yz+C$
Work Step by Step
The vector field $F$ will be conservative if and only if $curl F=0$
Let us consider $F=P i+Q j+R k$
Then, we have $curl F=[R_y-Q_z]i+[P_z-R_z]j+[Q_x-P_y]k$
Now, $curl F=[(e^{x}\cos yz-yze^{x}\sin yz)-(e^{x} \cos yz-yze^{x} \sin yz)]i+[(ye^{x} \cos yz-ye^{x} \cos yz)]j+[(ze^{x} \cos yz-ze^{x} \cos yz)-k=0$
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 $f_y=z e^{x} \cos yz$
Further, $f(x,y,z)=e^{x} \sin yz+h(z)$
Thus, we have $f_z=y e^{x} \cos yz+h'(z)$
and $h'(z)=0$
Hence, we get $f(x,y,z)=e^{x} \sin yz+C$
where, $C$ is a constant of proportionality.