[tex]\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k[/tex]
We want to find [tex]f(x,y,z)[/tex] such that [tex]\nabla f=\vec F[/tex]. This means
[tex]\dfrac{\partial f}{\partial x}=x^2+y[/tex]
[tex]\dfrac{\partial f}{\partial y}=y^2+x[/tex]
[tex]\dfrac{\partial f}{\partial z}=ze^z[/tex]
Integrating both sides of the latter equation with respect to [tex]z[/tex] tells us
[tex]f(x,y,z)=e^z(z-1)+g(x,y)[/tex]
and differentiating with respect to [tex]x[/tex] gives
[tex]x^2+y=\dfrac{\partial g}{\partial x}[/tex]
Integrating both sides with respect to [tex]x[/tex] gives
[tex]g(x,y)=\dfrac{x^3}3+xy+h(y)[/tex]
Then
[tex]f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)[/tex]
and differentiating both sides with respect to [tex]y[/tex] gives
[tex]y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C[/tex]
So the scalar potential function is
[tex]\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}[/tex]
By the fundamental theorem of calculus, the work done by [tex]\vec F[/tex] along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it [tex]L[/tex]) in part (a) is
[tex]\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}[/tex]
and [tex]\vec F[/tex] does the same amount of work over both of the other paths.
In part (b), I don't know what is meant by "df/dt for F"...
In part (c), you're asked to find the work over the 2 parts (call them [tex]L_1[/tex] and [tex]L_2[/tex]) of the given path. Using the fundamental theorem makes this trivial:
[tex]\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3[/tex]
[tex]\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4[/tex]
