Let $X$ Bananch space and $f \colon X \to \mathbb R$, $A$ open in $X$. $f\in \mathcal C^1(A)$ if it Frechet differentiable for all $x \in A$ and its Frechet derivative $Df \colon X \to \mathcal L (X, \mathbb R)=X'$ is continuous in the dual norm.
Now consider the case $X = \mathbb R^n$. Why do we say that $f \in \mathcal C^1(A)$ is there exist all partial derivatives and they are continuous? This is in principle a weaker condition wrt to the condition for a general Banach. Are they equivalent on $\mathbb R^n$?
(We assume we are in $\mathbb{R^{n}}$). If all partial derivatives exist and are continuous, then is Frechet differentiable (since $\bigtriangledown f(x)$ exists and it is continuous) and also $C^{1}$.
But the inverse statement is not true. Consider:
$f(x,y)=(x^{2}+y^{2})\sin((x^{2}+y^{2})^{-1/2})$ if $(x,y)\neq\,(0,0)$
and $f(0,0)=0$.
This is Frechet differentiable but it does not have continuous partial derivatives at $(0,0)$.
But if we assume that an $f$ is Frechet differentiable and also in $C^{1}$, i.e. $\bigtriangledown f(x)$ exists and it is continuous, then clearly all partial derivatives exist and are continuous.