Find a sufficient condition for a point $(x_0,y_0,z_0)$ in the intersection C of surfaces $x^2+y^2+z^2=1$ and $x^3+y^3+z^3=0$ to have a neighborhood $U$ with $U\cap C$ parametrized by the x coordinate as a curve, that is, $U \cap C = \{(x,f(x)): x \in (a,b)\}$ for some interval $(a,b)$ and some function $f:(a,b) \rightarrow \mathbb{R}^2$.
Attempt: Not sure how to start. I think I have to apply the Implicit Function Theorem. The intersection of surfaces would be given by $x^3-x^2+y^3-y^2+z^3-z^2+1=0$. But I am stuck. How to proceed from here?
This is the implicit function theorem, as you suspected. Define the function $$f(x,y,z) = \pmatrix{x^2+y^2+z^2-1\\ x^3+y^3+z^3}$$ so that $$f(x_0,y_0,z_0)=\pmatrix{0\\0}$$
The Jacobian of $f$ at $(x_0,y_0,z_0)$ is $$Df(x_0,y_0,z_0)= \left( \begin{array}{c|cc} 2x_0&2y_0&2z_0\\3x_0^2&3y_0^2&3z_0^2 \end{array} \right) $$
where the vertical line indicates that since we want to express $y,z$ as functions of $x$, we will consider the partial derivatives with respect to $y$ and $z$, that is, the last two columns of the matrix. The condition is $$\left |\matrix{2y_0&2z_0\\3y_0^2&3z_0^2} \right|\ne0$$ that is $$\boxed{y_0z_0(z_0-y_0)\ne0}$$