I am achieving the wrong answer, but I am unsure where I went wrong about it. I am given a conservative vector field and the line $C$: $$F = \langle \frac{4x}{y^2+1}, -\frac{4y(x^2+1)}{(y^2+1)^2} \rangle, C \textrm{ is parameterized by } x=t^3-1, y=t^6-t, 0 \le t \le 1 .$$
Since I know it is conservative, I find the potential function $f$: $$\frac{4x^2+2}{y^2+1} + c$$
Then I just apply FTLI, $f(x(1), y(1)) - f(x(0), y(0))$ by recognizing $t$ is in between $0$ and $1$ therefore its initial point must be $(-1,0)$ and its terminal point $(0,0)$. So I get $2 - 6 = -4$, but $-4$ is deemed wrong.
I am not sure when I went wrong in this, I double checked my integrals for $P$ and $Q$ when getting the potential function, did I go wrong in determining the initial and terminal points?
Your pontential is not correct, it should be $$f(x,y)=\frac{2x^2+2}{y^2+1}.$$ Also, $$f((x(1),y(1))=f(0,0)=2,$$ and $$f(x(0),y(0))=f(-1,0)=4,$$ so the correct answer is $2-4=-2.$