Key idea: $x^2$ is always greater than or equal to 0
This means $-x^2$ is always less than or equal to 0
And this means $-x^2 - 3$ is always NEGATIVE
In other words, the denominator of $y$ is most definitely NEGATIVE

From here, there are two possible cases:
case i: If the numerator of $y$ is NEGATIVE, then y is POSITIVE
case ii: If the numerator of $y$ is POSITIVE, then y is NEGATIVE

Since we're trying to MAXIMIZE the value of y, we need the denominator to be negative (which would make y POSITIVE)
However, the denominator, |3x - 5|, will always be greater than or equal to 0.
Since we can't make the numerator NEGATIVE, the only way to maximize the value of y is make the numerator ZERO
In other words: $|3x - 5|=0$
Which means $3x -5=0$
So $3x =5$
Ergo $x = \frac{5}{3}$

Answer: E

Cheers,
Brent