Evaluate using Stokes' Theorem

Hint: Since $x^2+y^2=1$ you have that $$x=\cos t\quad y=\sin t\quad 0\leq t\leq2\pi.$$ Thus, a parametrization for the curve $C$ is

$$x=\cos t\quad y=\sin t\quad z=1-\cos t-\sin t\quad 0\leq t\leq2\pi.$$

Let $\mathbf{F}=\left(-y^3,x^3,z^3\right).$ So we have \begin{align*} \oint_{C} -y^3dx+x^3dy+z^3dz&=\oint_C\mathbf{F}\cdot d\mathbf{r} \\ &=\iint_S(\nabla\times\mathbf{F})\cdot dS \\ &=\iint_S(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}}\,dA \\ &=\sqrt{3}\iint_S\left(x^2+y^2\right)\,dA. \end{align*} My hunch is that we can consider the projection into the $xy$ plane of the surface $S$ for this problem, because the integrand $\nabla\times\mathbf{F}$ only has a $z$ component. If that is so, we will want to switch to polar coordinates: \begin{align*} \oint&=\sqrt{3}\int_0^{2\pi}\int_0^1\left(x^2+y^2\right)\,r\,dr\,d\theta \\ &=2\sqrt{3}\,\pi\int_0^1 r^3\,dr \\ &=\frac{\sqrt{3}\,\pi}{2}. \end{align*} As we can see from this wiki, to adjust for the projection, we need $$A_{\text{proj}}=\cos(\beta) A, $$ since the angle is constant and pulls out of the integral. We have computed the projected area, so we must compensate by dividing by $\cos(\beta),$ we can calculate via the dot product formula: $$\frac{1}{\sqrt{3}}(1,1,1)\cdot(0,0,1)=\cos(\beta). $$ This means the final result is $$\frac{\sqrt{3}\,\pi}{2}\div\frac{1}{\sqrt{3}}=\frac{3\pi}{2}. $$

Now the question is, was the projection justified? Can we verify by, say, computing the original line integral? As suggested by DiegoMath in his answer, we can parametrize the curve $C$ as \begin{align*} x&=\cos(t) \\ y&=\sin(t) \\ z&=1-\cos(t)-\sin(t),\\ 0&\le t\le 2\pi. \end{align*} Then we have \begin{align*} \mathbf{r}(t)&=(\cos(t), \sin(t), 1-\cos(t)-\sin(t))\\ \dot{\mathbf{r}}(t)&=(-\sin(t), \cos(t), \sin(t)-\cos(t)) \\ \mathbf{F}(t)&=(-\sin^3(t),\cos^3(t),(1-\cos(t)-\sin(t))^3) \\ \mathbf{F}\cdot\dot{\mathbf{r}}&=\sin^4(t)+\cos^4(t)+(\sin(t)-\cos(t))(1-\cos(t)-\sin(t))^3 \\ \oint_C\mathbf{F}\cdot d\mathbf{r}&=\oint_C\mathbf{F}\cdot \dot{\mathbf{r}}(t)\,dt \\ &=\int_0^{2\pi}\left[\sin^4(t)+\cos^4(t)+(\sin(t)-\cos(t))(1-\cos(t)-\sin(t))^3\right]dt \\ &=\frac{3\pi}{2}, \end{align*} which is the answer we had above.