This one should be a quickie:
if we are calculating $\oint xdy+ydx$ around the curve $x^4 + y^4 = 1$, I can spot quickly that $\oint xdy+ydx = \oint df$ where $f = xy$. How would I then evaluate $\oint df$? Is it simple, or am I better off with a parameterisation approach ($x^2 = \mathrm{cos} \ t$ and so on)?
Your form $\omega=ydx+xdy$ is defined in $\Omega=\mathbb R^2$, which is contractible, hence it's simply connected (the path $\gamma\equiv x^4+y^4=1$ is closed and continuous).
We can now solve the exercise using the fact that a form in $\Omega$ is closed $\iff$ the form is exact, so we observe that $$\dfrac{\partial}{\partial_y}y=1=\dfrac{\partial}{\partial_x}x\overset{DF}{\implies}\omega\text{ is closed}\implies\omega\text{ is exact.}$$