=>

If x = 0, then f(0,y) = 20y / 2y = 10.
If x = y, then f(y,y) = 10y / 5y + 20y / 5y = 2 + 4 = 6.
Thus, if 0 < x < y, we have 6 < f(x,y) < 10.
The only possible value of f(x,y) from among the choices is 8.

Therefore, A is the answer.
Answer: A

The function above has 2 arguments x and y. This makes it difficult to find f(x,y) but using 0(x,y).
$$if\ x=0,\ then$$
$$f\left(0,y\right)=\frac{10\left(0\right)}{2\left(0\right)+3y}+\frac{20y}{3\left(0\right)+2y}$$
$$f\left(0,y\right)=\frac{0}{3y}+\frac{20y}{2y}$$
$$f\left(0,y\right)=\frac{20y}{2y}=10$$

$$If\ x=y,\ then;$$
$$f\left(y,y\right)=f\left(x,x\right)=\frac{10y}{2y+3y}+\frac{20y}{3y+2y}$$
$$f\left(y,y\right)=f\left(x,x\right)=\frac{10y}{5y}+\frac{20y}{5y}$$
$$\frac{30y}{5y}=6$$

From 0<x<y, x is the range between 0 and y excluding 0 and y. So, f(x,y) is the range between 6 and 10 excluding 6 and 10. This will be 6<f(x,y)<10.
The only option that satisfies this inequality is option A => 8

Therefore, 6 < f(x,y) < 10 where f(x,y) = 8. Hence, 6 < 8 < 10