[tex]\mathbf f(x,y,z)=yze^{xz}\,\mathbf i+e^{xz}\,\mathbf j+xye^{xz}\,\mathbf k[/tex]
We're looking for a scalar function [tex]f(x,y,z)[/tex] such that its gradient is equal to the given vector-valued function [tex]\mathbf f(x,y,z)[/tex]:
[tex]\nabla f(x,y,z)=\mathbf f(x,y,z)[/tex]
That is, we require
[tex]\dfrac{\partial f}{\partial x}=yze^{xz}[/tex]
[tex]\dfrac{\partial f}{\partial y}=e^{xz}[/tex]
[tex]\dfrac{\partial f}{\partial z}=xye^{xz}[/tex]
Integrating the first equation with respect to [tex]x[/tex] gives
[tex]f(x,y,z)=\dfrac{yz}xe^{xz}+g(y,z)[/tex]
Differentiating with respect to [tex]y[/tex] gives
[tex]\dfrac{\partial f}{\partial y}=e^{xz}=\dfrac zxe^{xz}+\dfrac{\partial g}{\partial y}[/tex]
but from this step we can see that it's not possible for [tex]g[/tex] to be a function of [tex]y[/tex] and [tex]z[/tex], independent of [tex]x[/tex]. So there is no such function [tex]f(x,y,z)[/tex].
