Define:
$$\vec{y}_:=(y_1,y_2,\cdots,y_n), ~~~ \vec{f}_:=(f_1,f_2,\cdots,f_n),$$
$\vec{\lambda}_:=(\lambda_1,\lambda_2,\cdots,\lambda_3)$ are the parameters of the differential equations.
$$d\vec{y}/dx=\vec{f}(x,\vec{y},\vec{\lambda}),~~~ \vec{y_0}=\vec{f}(x_0)$$ is a group of differential equations which mean \begin{align} dy_1/dx&=f_1(x,y_1,y_2,\cdots,y_n,\vec{\lambda}),\\ dy_2/dx&=f_2(x,y_1,y_2,\cdots,y_n,\vec{\lambda}),\\ \cdots,\\ dy_n/dx&=f_n(x,y_1,y_2,\cdots,y_n,\vec{\lambda}) \end{align} and satisfying the initial condition $y_{01}=f_1(x_0),\cdots,y_{0n}=f_n(x_0)$. Suppose $x\in D$, $\vec{y} \in \Omega$, $\vec{\lambda}\in \Lambda $. If $\vec{f} \in C^k(D \times\Omega\times\Lambda)$ (where $C^k $ means the derivative of $k$ order is still continuous in the definition domain), then we can have $x,x_0 \in C^k(D)$, $\vec{y_0} \in C^k(\Omega)$, $\lambda \in C^k(\Lambda)$.
When $k=0,1$, it's also not trivial, but it has been proved by me.
I have an intuition that the proof may involve the variational equation (matrix).
It's a really interesting problem, have a challenge or show your ideas！