[tex]\vec F(x,y,z)=(e^x+y)(1,1,z)[/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]f_x=e^x+y[/tex]
[tex]f_y=e^x+y[/tex]
[tex]f_z=(e^x+y)z[/tex]
Integrating both sides of the first equation wrt [tex]x[/tex] gives
[tex]f(x,y,z)=e^x+xy+g(y,z)[/tex]
Differentiating both sides of this wrt [tex]y[/tex] gives
[tex]f_y=x+g_y=1\implies g_y=1-x\implies g(y,z)=y-xy+h(z)[/tex]
but we assumed [tex]g[/tex] was a function of [tex]y[/tex] and [tex]z[/tex], independent of [tex]x[/tex]. So there is no such [tex]f[/tex] and [tex]\vec F[/tex] is not conservative.
To find the work, first parameterize the path (call it [tex]C[/tex]) by
[tex]\vec r(t)=(1-t)(0,0,0)+t(-4,5,-5)=(-4t,5t,-5t)[/tex]
for [tex]0\le t\le1[/tex]. Then
[tex]\vec r'(t)=(-4,5,-5)[/tex]
and the work is given by the line integral,
[tex]W=\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_0^1(e^{-4t}+5t,e^{-4t}+5t,-(e^{-4t}+5t)5t)\cdot(-4,5,-5)\,\mathrm dt[/tex]
[tex]W=\displaystyle\int_0^1(125t^2+5t+(25t+1)e^{-4t})\,\mathrm dt=\boxed{\frac{2207}{48}-\frac{129}{16e^4}}[/tex]
