For a matrix A, is there a function f(A) such that $$\frac{df_{ij}}{dA_{kl}}=f_i^k (A) \delta^l_j$$ ?
I am specifically interested in the case of a symmetric matrix $A$ and function f of rank 2.
Prior research: I tried $e^A$ but it does not have this property. The ij element of $e^A$ is $$e^A_{ij} = \sum_{n=0}^\infty \frac{1}{n!} A_i^{k_1} ... A_{k_{n-1}j} ~.$$ Each matrix element can be seen as an independent variable, so the derivative toward $A^{kl}$ is $$e^A_{ij} = \sum_{n=0}^\infty \frac{1}{n!} \sum_{p=0}^{n-1} A^p_{ik} A^{n-1-p}_{lj} ~.$$ This does not have the desired form.
The $\delta_{jl}$ on the right means that $f_{ij}(A)$ depends on $A_{kl}$ only if $j=l$, that is $f_{ij}(A) = g_{ij}(A_{1j}, \dots, A_{nj})$ for some scalar functions $g_{ij}(x_1, \dots, x_n)$.
Plugging $g_{ij}$ into the equation gives $$ \frac{\partial f_{ij}}{\partial A_{kl}} = \delta_{jl} \frac{\partial g_{ij}}{\partial x_k}(A_{1j}, \dots, A_{nj}). $$
On the other hand, the right hand side is $f_{ik}(A) \delta_{jl}$. We've arived to an equation for $g$: $$ \frac{\partial g_{ij}}{\partial x_k}(A_{1j}, \dots, A_{mj}) = g_{ik}(A_{1k}, \dots, A_{mk}), \quad i, j, k = 1,\dots,n \tag{*} $$ When $j \neq k$ the expression on the left depends on $A_{1j}, \dots, A_{mj}$, that is on the $j$-th column of the matrix, but the expression on the right depends on $k$-th column of the matrix. The only possibility of equality is that $g_{ij}$ completely don't depend on $A$ elements at all!
Since $g_{ij}$ don't depend on $A_{kl}$, the derivative on the left of (*) is zero, so should be the right side $g_{ik}$.
Finally, the only possible solution is $f(A) = 0$.