Work on vector field $= 0$

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I am doing the problem

Calculate the work done in a moving particle along the given path

$F=<yze^x+xyze^x,xze^x,xye^x>$ along $C: r(t)= <2\cos t,3\sin t,0>$

$1≤t≤2$

I found that this field was conservative, so I was able to set up a potential function

$f=xyze^x+C$

From there to calculate work I did

$f_{end}-f_{start} $

This gives me zero though because of the z component. Is this correct or am I doing something wrong? Thanks in advance.

Edit: Showing requested work

$r(t)= <2\cos t,3\sin t,0>$ $1≤t≤2$

$r(1)= <2\cos(1),3\sin (1),0>$

$r(2)= <2\cos (2),3\sin (2),0>$

$f_{end}-f_{start} $

$ 0 - 0 = 0 $