c(t) = (cos^3(t),sin^3(t)) t goes from 0 to 2pi
F(x,y) = xi + yj
I know the equations but something tells me if i do the gradient it equals 0. The problem I'm having is how to do and find the gradient.
Why gradient? That is the opposite of the integral. I presume that you mean you are to integrate $\displaystyle \int_C F(x,y)\cdot d\vec{s}= \int_C xdx+ ydy$. If $\displaystyle C(t)= cos^3(t)\vec{i}+ sin^3(t)\vec{j}$ then $\displaystyle d\vec{s}= -3sin(t)cos^2(t)dt\vec{i}+ 3sin^2(t)cos(t)dt\vec{j}$ or $\displaystyle dx= -3sin(t)cos^2(t)dt$ and $\displaystyle dy= 3sin^2(t)cos(t)dt$
$\displaystyle F(x,y)= cos^3(t)\vec{i}+ sin^3(t)\vec{j}$ so you integral is
$\displaystyle \int_0^{2\pi} (-3sin(t)cos^5(t)+ 3sin^5(t)cos(t))dt$
Since that involves only odd powers of sine and cosine, it should be easy to integrate.