[tex]\vec F[/tex] is conservative if we can find a scalar function [tex]f[/tex] such that [tex]\nabla f=\vec F[/tex]. This would require
[tex]\dfrac{\partial f}{\partial x}=5y^2z^3[/tex]
[tex]\dfrac{\partial f}{\partial y}=10xyz^3[/tex]
[tex]\dfrac{\partial f}{\partial z}=15xy^2z^2[/tex]
Integrate both sides of the first PDE with respect to [tex]x[/tex]:
[tex]f(x,y,z)=5xy^2z^3+g(y,z)[/tex] (*)
Differentiate both sides of (*) with respect to [tex]y[/tex]:
[tex]\dfrac{\partial f}{\partial y}=10xyz^3=10xyz^3+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)[/tex]
Differentiate both sides of (*) with respect to [tex]z[/tex]:
[tex]\dfrac{\partial f}{\partial z}=15xy^2z^2=15xy^2z^2+\dfrac{\mathrm dh}{\mathrm dz}\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C[/tex]
So we have
[tex]f(x,y,z)=5xy^2z^3+C[/tex]
and so [tex]\vec F[/tex] is indeed conservative.
