Let $E_i$ be a normed $\mathbb R$-vector space.
From Schwarz's theorem we know that if $\Omega\subseteq E_1$ is open and $f:\Omega\to E_2$ is twice differentiable at $x\in\Omega$, then ${\rm D}^2f(x)$ is symmetric; i.e. $${\rm D}^2f(x)(h,h')={\rm D}^2f(x)(h',h)\tag1$$ for all $h,h'\in E_1$.
Now suppose $\Omega\subseteq E_1\times E_2$ and $f:\Omega\to E_3$ is twice differentiable at $x\in\Omega$. Can we deduce that $$({\rm D}_2({\rm D}_1f)(x)h_2)h_1'=({\rm D}_1({\rm D}_2f)(x)h_1)h_2'\tag2$$ for all $h,h'\in E_1\times E_2$, where ${\rm D}_i$ denotes the Fréchet derivative with respect to the $i$th argument, from Schwarz's theorem?