Solution:
Assume that x = y = 0, then according to the relationship given in question:f(x+y) = f(x)+f(y)
f(0+0) = f(0)+f(0)
f(0) = 0
Now let us consider y = −x and check the nature of function.
f(x+y) = f(x)+f(y)
f(x+(−x)) = f(x)+f(−x)
f(0) = f(x)+f(−x)
0 = f(x)+f(−x)
f(x) = −f(−x)
As the function satisfies the relation f(x) = −f(−x). So, the function f(x) is an odd function.
An odd function is never an even function. So, the function f(x) is not an even function.
