I an unable to answer this question. I got a potential function of $V(x,y,z) =yzx^2$. I also know I must plug in $c(0)$ and $c(8)$. I got $c(0)=(0,0,1)$ and $c(8)=(64,0,e^{48})$. I then plug these numbers in to my potential function and subtract the lower number from the larger number. When plugged into the potential function I receive the answer of $0$. This is not the answer. I would appreciate any help whether it be an answer, steps, or advice on where I made a mistake.
Well, if $V(x,y,z) = x^2yz$, we have: $$\mathrm{d}V = 2xyz~\mathrm{d}x + x^2z~\mathrm{d}y + x^2y~\mathrm{d}z$$ so your potential is right. Also, $c(0) = (0,0,1)$, ok, and $c(8) = (64, 0, e^{48})$. So far, so good. Finally, $$\int_C 2xyz~\mathrm{d}x + x^2z~\mathrm{d}y + x^2y~\mathrm{d}z = V(c(8)) - V(c(0)) = 0$$
Since you said this is not the answer, let's do it directly. We have $$\begin{align} \mathrm{d}x &= 2t~\mathrm{d}t \\ \mathrm{d}y &= \frac{\pi}{4}\cos \left(\frac{\pi t}{4}\right)~\mathrm{d}t \\ \mathrm{d}z &= (2t - 2)e^{t^2 - 2t}~\mathrm{d}t \end{align}$$ and hence $$\int_C 2xyz~\mathrm{d}x + x^2z~\mathrm{d}y + x^2y~\mathrm{d}z = \int_0^8 4t^3\sin \left(\frac{\pi t}{4}\right)e^{t^2-2t} + \frac{\pi}{4}\cos\left(\frac{\pi t}{4}\right)t^4e^{t^2 - 2t} + (2t - 2)e^{t^2-2t}t^4\sin\left(\frac{\pi t}{4}\right)~\mathrm{d}t$$ I believe there's an mistake in the answer the text gives, because Wolfram Alpha also returns zero to this integral.