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 $