I want to prove that for $A \subseteq \mathbb R^n$ open set and $f:A\to \mathbb R$
if for all $i,j = 1,\dots, n$ $$\frac{\partial ^2f}{\partial x_i\partial x_j}(x)$$ is continous on $A$, then $$\frac{\partial f}{\partial x_j}(x)$$ is also continuous on $A$ for $j=1,\dots,n$
However I don't want to use the fact that if all second partial derivatives exist and are continuous, then $f \in C^2(A)$.
From the definition $\frac{\partial^2 f}{\partial x_i\partial x_j}:=\frac{\partial }{\partial x_i}\left[\frac{\partial f}{\partial x_j}\right]$, you are saying that $\frac{\partial f}{\partial x_j}$ has continuous partial derivatives. Therefore it is a differentiable function.