f(x+y)=f(x)+f(y); let x=5 and y=0, then f(x+y)=2 and f(x)+f(y)=f(5)+f(0)=2+f(0)=2, making f(0)=0.
The point (0,0) therefore lies on f. The slope at this point is f'(0)=3.
If f were a quadratic function or higher degree then f(x+y) could not be the sum of f(x) and f(y). For example, if f(x)=x2, then f(y)=y2, (the argument of f is simply a placeholder in the definition of the function) so f(x)+f(y)=x2+y2, and f(x+y)=(x+y)2=x2+2xy+y2, which contains an additional term. From this I conclude that f is a linear function. However, f(5)=2, so a linear function joining (0,0) to (5,2) would have a constant slope of ⅖. Since the slope at (0,0) is 3, the function must fall back on itself linearly (perhaps a function containing absolute value) would do this. I conclude that the slope at (5,2)=f'(5)=-3.