Answer
Conservative; $f(x,y,z)=xe^{yz}+K$
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 $curl F=[(xe^{yz}-xyze^{yz})-(xe^{yz}-xyze^{yz})]i+[(ye^{yz}-ye^{yz})]j+[(ze^{yz}-ze^{yz})-k=0$
Thus, the vector field $F$ is conservative.
$f(x,y,z)=xe^{yz}+g(y,z)$
$f_y=xze^{yz}+g_y \implies g'(y)=0$
Thus, $g_y=h(z)$ and $f_y=xze^{yz}$
$g(y,z)=y \sin z+h(z)$
Now, $f(x,y,z)=xe^{yz}+h(z)$
Thus, $f_z=xye^{yz}+h'(z)$
$ \implies h'(z)=0$
Hence, $f(x,y,z)=xe^{yz}+K$
