Prove that every point $(x_0,y_0,z_0), 0<y_0<z_0,$ which is a solution to the system $$\begin{cases}x^2+y^2+z^2=2\\x^3+y^3+z^3=1\end{cases}$$ lies on a smooth curve.
I interpreted smooth curve to mean the trace/image of a parametrized curve $\gamma:I\subseteq\Bbb R\to\Bbb R^3,\gamma(t)=(\gamma_1(t),\gamma_2(t),\gamma_3(t)),\gamma_i:\Bbb R\to\Bbb R.$
I thought of applying the Inverse Function Theorem in this manner, so suppose $(x_0,y_0,z_0)$ is a solution and let $$ \begin{aligned}f_1(x,y,z)&=x\\f_2(x,y,z)&=x^2+y^2+z^2-2\\f_3(x,y,z)&=x^3+y^3+z^3-1\end{aligned}$$ Hence $f\in C^1(\Bbb R^3;\Bbb R^3),\quad f(x_0,y_0,z_0)=(x_0,0,0)$ and $$\begin{aligned}J_f(x,y,z)&=\det\nabla f(x,y,z)\\&=\begin{vmatrix}1&0&0\\2x&2y&2z\\3x^2&3y^2&3z^3\end{vmatrix}\\&=6yz(z-y),\end{aligned}$$ so, as $0<y_0<z_0, J_f(x_0,y_0,z_0)\ne 0$. By the implicit function theorem, there is an open neighbourhood $U$ of $(x_0,y_0,z_0)$ and $V$ of $(x_0,0,0)$ s. t. $f:U\to V$ is a diffeomorphism, that is, $f$ has an inverse $f^{-1}:V\to U$ of class $C^1$ and now we can define a parametrized curve $\gamma(t)=f^{-1}(t,0,0)=(t,y(t),z(t)).$
$\gamma$ is of the class $C^1$ around $x_0$ because $f^{-1}$ is of class $C^1$ around $(x_0,0,0).$
Question: Is there anything I have misunderstood in the linked answer and or anything to improve?