Let's test a set of values: $y = 3$ and $z = 1$

If the given condition is true, then it must be the case that $f(3 − 1) = f(3) − f(1)$
When we simplify the first function, we get: $f(2) = f(3) − f(1)$

So for each answer choice, let's test whether $f(2) = f(3) − f(1)$

A. $f(x) = x^2$
$f(2) = f(3) − f(1)$ becomes $2^2 = 3^2 − 1^2$
Simplify: $4 = 9 − 1$
NOT TRUE.
Eliminate A

B. $f(x) = x + (x − 1)^2$
$f(2) = f(3) − f(1)$ becomes $2 + (2-1)^2 = (3 + (3-1)^2) − (1 + (2-1)^2)$
Simplify: $3 = 7 − 2$
NOT TRUE.
Eliminate B

C. $f(x)=x−1$
$f(2) = f(3) − f(1)$ becomes $2 - 1 = (3 - 1) − (1 - 1)$
Simplify: $1 = 2 − 0$
NOT TRUE.
Eliminate C

D. $f(x)=\frac{5}{x}$
$f(2) = f(3) − f(1)$ becomes $\frac{5}{2} = \frac{5}{3} - \frac{5}{1}$
Simplify: $\frac{5}{2} = -\frac{10}{3}$
NOT TRUE.
Eliminate D

By the process of elimination, the correct answer must be E.
But let's check to make sure.

E. $f(x)=\frac{x}{5}$
$f(2) = f(3) − f(1)$ becomes $\frac{2}{5} = \frac{3}{5} - \frac{1}{5}$
WORKS!

Answer: E