Limits over a functor commute It is given here https://ncatlab.org/nlab/show/limits+commute+with+limits
that the if a category admits limits of shapes $D$ and $D'$, then it admits limits of shape $D\times D'$, and furthermore, we have the isomorphism
$\lim F \simeq \lim _{\mathscr{D}}\left(\lim _{\mathscr{D'}}  F_{\mathscr{D}}\right) \simeq \lim _{\mathscr{D}^{\prime}}\left(\lim _{\mathscr{D}} F_{\mathscr{D}^{\prime}}\right)$
It is cavalierly mentioned that this is a consequence of "right adjoints preserves limits".
Attempt:
Put $\Xi_{D}:Fct(C\times D,\mathbb{A})\rightarrow Fct(D,Fct(C,\mathbb{A}))$ as the isomorphism of categories. For shorthand, denote $(L\alpha,\lambda^\alpha)$ as the limit of $\alpha$. $\lim_{C,\mathbb{A}}:Fct(C,\mathbb{A})\rightarrow\mathbb{A}$ is the limit functor and is therefore right adjoint.
$\Xi_{D}\alpha:D\rightarrow Fct(C,\mathbb{A})$ is a functor, with limit $(\text{L}_{D}\Xi_{D}\alpha,\lambda^{\Xi_{D}\alpha})$ (Question: does this limit actually exist?). So we can compose it with $\lim_{C,\mathbb{A}}$
Therefore we have that $(\lim_{C,\mathbb{A}}\text{L}_{D}\Xi_{D}\alpha,\lim_{C,\mathbb{A}}\lambda^{\Xi_{D}\alpha})$ is a limit of the functor $\lim_{C,\mathbb{A}}\circ\Xi_{D}\alpha$ and is therefore isomorphic to $\text{L}_{C}\text{L}_{D}\Xi_{D}\alpha$
I am pretty much stuck and out of ideas beyond this point.
 A: I can't really read what you wrote because of your notations; but here are the two main ideas :
1- The limit functor $Fun(C,A) \to A$ is right adjoint to the diagonal functor (you noted that one in your question)
2- If $A$ has all limits of type $D$, then $Fun(C,A)$ does too, and they are computed pointwise, that is, each evaluation functor $ev_c : Fun(C,A) \to A$ preserves limits (and they jointly reflect them).
Once you have this, it becomes pretty easy : Suppose you have your functor $D\times C\to A$. You may see it as a functor $D\to Fun(C,A)$.
If you compute the limit of that, and then apply the limit functor, you get a limit in $A$, which may be denoted $\lim_C \lim_D F$ (because limits in $Fun(C,A)$ are computed pointwise). But the $\lim_C$ functor preserves limits (because it is a right adjoint), so this is also $\lim_D\lim_C F$
To compare them with $\lim_{D\times C}$, you should indeed use the equivalence $Fun(D\times C,A)\simeq Fun(D,Fun(C,A))$ and see what the diagonal functor $A\to Fun(D\times C,A)$ becomes under this equivalence (and so, by unicity of right adjoints, you should be able to compare the functors you want to compare)
Just a remark : these proofs use strong assumptions to get conceptual proofs, but there are also "local" statements that can't be proved so conceptually, although the proof is very simple. In particular, you don't need to assume all limits of type $C$ and all limits of type $D$ exist, you only need the ones that appear in the statements (the ones you "obviously" need).
It's then just a matter of checking by hand the universal property, which is not so bad.
