Answer
$f(x,y,z)=xy^2z^3+C$
Work Step by Step
The vector field $F$ will be conservative if and only if $curl F=0$
Let us consider $F=P i+Q j+R k$
$curl F=[R_y-Q_z]i+[P_z-R_z]j+[Q_x-P_y]k$
Here, we have $curl F=(6x yz^2 -6xyz^2)+(3y^2z^2-3y^2z^2)+(2yz^3-2yz^3)=0$
Therefore, the vector field $F$ is conservative.
Also, $f(x,y,z)=xy^2z^3+g(y,z)$
$ \implies g'(y)=0$ and $F_y=2xyz^3$
Further, $f(x,y,z)=xy^2z^3+h(z) \implies h'(z)=0$
and $F_z=3xy^2z^2$
Thus, $f(x,y,z)=xy^2z^3+C$ [ C is constant of proportionality]
