Let $E_1$, $E_2$ and $F$ be normed $\mathbb R$-vector spaces, $\Omega\subseteq E_1\times E_2$ and $f:\Omega\to F$ be Fréchet differentiable at $x\in\Omega^\circ$.
Are we able to show that $${\rm D}f(x)y={\rm D}_1f(x)y_1+{\rm D}_2f(x)y_2=:Ly\;\;\;\text{for all }y\in E_1\times E_2?\tag1$$
We know that $$\frac{\left\|f(x+y)-f(x)-{\rm D}f(x)y\right\|_F}{\left\|y\right\|_{E_1\times E_2}}\xrightarrow{y\to0}0\tag2$$ and $$\frac{\left\|f(x_1+y_1,x_2)-f(x)-{\rm D}_1f(x)y_1\right\|_F}{\left\|y_1\right\|_{E_1}}\xrightarrow{y_1\to0}0\tag2,$$$$\frac{\left\|f(x_1,x_2+y_2)-f(x)-{\rm D}_2f(x)y_2\right\|_F}{\left\|y_2\right\|_{E_2}}\xrightarrow{y_2\to0}0.\tag3$$
So, $$\frac{\left\|f(x+y)-f(x)-Ly\right\|_F}{\left\|y\right\|_{E_1\times E_2}}\le\frac{\left\|f(x+y)-f(x_1+y_1,x_2)-f(x_1,x_2+y_2)+f(x)\right\|_F}{\left\|y\right\|_{E_1\times E_2}}+\frac{\left\|f(x_1+y_1,x_2)-f(x)-{\rm D}_1f(x)y_1\right\|_F}{\left\|y_1\right\|_{E_1}}+\frac{\left\|f(x_1,x_2+y_2)-f(x)-{\rm D}_2f(x)y_2\right\|_F}{\left\|y_2\right\|_{E_2}}.\tag4$$
The second and third term tend to $0$ by $(2)$ and $(3)$, but what can we do with the first term?