The partial derivative of the gradient function I feel ashamed asking this basic question but am still confused.

Given function $f \in C^2$, show that $g:=\text{grad} \, f$ is a $C^1$ function.

I understand we have to show that the 1st order partial derivatives of $g$ exist and are continuous, but since $g : U \to \mathbb{R}^n$ is a vector valued function.. How do I figure out the first order partial derivatives?
Edit: I tried to apply the chain rule and deduced that $$D_1(g\circ f)(x)=D_1 D_{1}^2f + \dots + D_1 D_{n}^2f$$ and since $f$ is a $C^2$ function, the second order derivatives of $f$ are continuous and thus the first order derivative of $g$ is also continuous. Am I on the right path?

Firstly, $g$ is differentiable precisely when its components are, so we can consider each component of $g$ individually. Let's consider the $i$th component
$$
g_i = \frac{\partial f}{\partial x_i}. 
$$
Now $f$ is twice differentiable, thus we can differentiate the right hand side, and by extension the left hand side. The expression for the derivative of $g_i$ is
$$
\frac{\partial g_i}{\partial x_j} = \frac{\partial^2 f}{\partial x_j \partial x_i} 
$$
