Answer
$f(x,y,z)=xe^{yz}+C$
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, $curl F=[(xe^{yz}-xyze^{yz})-(xe^{yz}-xyze^{yz})]i+[(ye^{yz}-ye^{yz})]j+[(ze^{yz}-ze^{yz})-k=0$
so, the vector field $F$ is conservative.
Consider $f(x,y,z)=xe^{yz}+g(y,z)$
and $f_y=xze^{yz}+g_y; g'(y)=0$
Here, $g(y,z)=y \sin z+h(z)$
This implies that $f(x,y,z)=xe^{yz}+h(z)$
Thus, $f_z=xye^{yz}+h'(z)$
and $h'(z)=0$
Hence, we have $f(x,y,z)=xe^{yz}+C$
where, $C$ is a constant of proportionality.