A three-dimensional vector field is conservative if it is also irrotational, i.e. its curl is [tex]\mathbf 0[/tex]. We have
[tex]\nabla\times\mathbf f(x,y,z)=-2e^{-x}\,\mathbf k[/tex]
so this vector field is not conservative.
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Another way of determining the same result: We want to find a scalar function [tex]f(x,y,z)[/tex] such that its gradient is equal to the given vector field, [tex]\mathbf f(x,y,z)[/tex]:
[tex]\nabla f(x,y,z)=\mathbf f(x,y,z)[/tex]
For this to happen, we need to satisfy
[tex]\begin{cases}f_x=ye^{-x}\\f_y=e^{-x}\\f_z=2z\end{cases}[/tex]
From the first equation, integrating with respect to [tex]x[/tex] yields
[tex]f_x=ye^{-x}\implies f(x,y,z)=-ye^{-x}+g(y,z)[/tex]
Note that [tex]g[/tex] *must* be a function of [tex]y,z[/tex] only.
Now differentiate with respect to [tex]y[/tex] and we have
[tex]f_y=-e^{-x}+g_y=e^{-x}\implies g_y=2e^{-x}\implies g(y,z)=2ye^{-x}+\cdots[/tex]
but this contradicts the assumption that [tex]g(y,z)[/tex] is independent of [tex]x[/tex]. So, the scalar potential function does not exist, and therefore the vector field is not conservative.