For $k=0,1,2,3,\dots$ and $X\in C^k(\mathbb{R}\times \mathbb{R}^n, \mathbb{R}^n)$ I want to show that any solution $(I,y)$ to $\dot{y}=X(t,y)$ satisfies that $y\in C^{k+1}(I,\mathbb{R}^n)$.
For $n=1$ it might not be so bad, since $X: \mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}$ and $y$ satisfies $\dot{y}(t) = X(t,y(t))$ for all $t\in I$, so on $I$ the derivative of $y$ is $X$, which is $k$ times continuous differentiable and $X$ itself is continuous, so it seems that $y$ is $k+1$ times continuously differentiable, but I fear that the higher dimension case becomes more complicated, since the derivatives become matrices... I don't quite know how to start, I considered induction in $n$ (induction in $k$ does not seem like the right approach).
Maybe one also needs to consider e.g. that $y^{(2)}(t) = \dot{X}(t, y(t))\dot{y}(t)$, which could be where the induction assumption is used.