Yeah, we are given a function on square matrices of a fixed size, call it $f,$ with three properties, square matrices $A,B$ and constant $c.$ So:
$$f(A + B ) = f(A) + f(B), $$
$$ f(AB) = f(BA), $$
$$ f(cA) = c f(A).$$
As Paul pointed out, the notation $e_{ij}$ means the matrix with a 1 at position $ij$ and 0 everywhere else.
There is some value for $f(e_{11}).$ E do not know what that is.
First,for some $i \neq 1,$ define
$$ S_i = e_{i1} + e_{1i} $$
The main thing is that
$$ S_i e_{11} S_i = e_{ii} $$
and $$ S_i^2 = I. $$
So
$$ f(e_{ii}) = f(S_i (e_{11} S_i)) = f( (e_{11} S_i) S_i) = f( e_{11} S_i^2) = f(e_{11}). $$
Next, with $i \neq j,$ we use
$$ e_{ii} e_{ij} = e_{ij} $$
while
$$ e_{ij} e_{ii} = 0, $$ the matrix of all 0's.
Begin with any $B,$
$$f(0) = f(0B) = 0 f(B) = 0.$$
Now, for any $i \neq j,$
$$ f(e_{ij}) = f(e_{ii} e_{ij}) = f(e_{ij} e_{ii}) = f(0) = 0. $$
Finally, if the entries of $A$ are $A_{ij},$ we have
$$ A = \sum_{i,j = 1}^n A_{ij} e_{ij}, $$ so
$$ f(A) = f(\sum_{i,j = 1}^n A_{ij} e_{ij}) = \sum_{i,j = 1}^n A_{ij} f(e_{ij}) = \sum_{i=1}^n A_{ii} f(e_{ii}) = \sum_{i=1}^n A_{ii} f(e_{11}) = f(e_{11}) \sum_{i=1}^n A_{ii} = f(e_{11}) \mbox{trace} A $$