Really simple question: If $X,Y,Z$ are Banach spaces, what's meant if we say that a function $f:X\times Y\to Z$ is continuously Fréchet differentiable in the first variable?
The Fréchet derivative in the first variable of $f$ is a function $X\times Y\to\mathfrak L(X,Z)$. So, does the continuity mean that this function is continuous in the first variable or does it mean that it is jointly continuous (with respect to the product topology on $X\times Y$)?
For each $y \in \mathrm{Y}$ one can define the partial mapping $x \mapsto f(x, y),$ denote this mappgin by $f( \cdot, y).$ That "$f$ is differentiable in the first variable" means that for every $y \in \mathrm{Y}$ the partial mapping $f( \cdot, y)$ is differentiable as a function $\mathrm{X} \to \mathrm{Z}.$ Denote the derivative of the partial mapping by $\mathbf{D}_1f(x, y),$ this is a a continuous linear mapping $\mathrm{X} \to \mathrm{Z};$ that is to say, for each $(x, y)$ the partial derivative $\mathbf{D}_1f(x, y)$ is an element of $\mathscr{L}(\mathrm{X}: \mathrm{Z}).$ However, we can clearly construct $(x, y) \mapsto \mathbf{D}_1f(x, y);$ if this mapping from $\mathrm{X} \times \mathrm{Y}$ into $\mathscr{L}(\mathrm{X}: \mathrm{Z})$ (the operator norm in the latter space) is continuous, then you obtain the definition of a "differentiable function in the first variable with continuity in both variables" which is how I would express "continuously Frechet differentiable in the first variable."