Useful property: $a-b = -1(b-a)$

So, $(x-y)^2=[(-1)(y-x)]^2$

$=(-1)^2(y-x)^2$

$=(1)(y-x)^2$

$=(y-x)(y-x)$

We're now ready to solve the question...

Given:
Quantity A: $y-x$
Quantity B: $(x-y)^2$

Replace Quantity B with its equivalent value:
Quantity A: $y-x$
Quantity B: $(y-x)(y-x)$

Since $0< x< y< 1$, we know that $y - x$ is POSITIVE, which means we can safely divide both quantities by $y - x$ to get:
Quantity A: $1$
Quantity B: $y-x$

From here, a nice way to determine which quantity is greater is to first add $x$ to both quantities to get:
Quantity A: $1+x$
Quantity B: $y$

Since x is positive, we know that Quantity A is greater than 1.
Conversely, we are told that y < 1.
So Quantity A must be greater.

Answer: A