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 $$ f(A) = f(\sum_{i,j = 1}^n A_{ij} e_{ij}) = \sum_{i,j = 1}^n A_{ij} f(e_{ij}) = \sum_{i=1}^n A_{ii} f(e_{ii}) = \sum_{i=1}^n A_{ii} f(e_{11}) = f(e_{11}) \sum_{i=1}^n A_{ii} = f(e_{11}) \mbox{trace} A $$