We could factor as follows:

x²y + 16xy + 64y = 0

y * (x² + 16x + 64) = 0

y * (x + 8)² = 0

Since y < 0, x must = -8. Since the equation is presumably a quadratic in x, we'd have one real root x. (I think the idea is that y is some known negative and that x is the only unknown, but the question is poorly phrased.)

The value of y is irrelevant as it doesn't affect the rest (x^2+16x+64) which must equal zero anyway (as anything times zero must be zero).
Therefore the question can be simplified to
How many real roots does the equation x^2+16x+64=0 ?

For any quadratic equation in the form ax^2+bx+c=0, we can determine the number of roots from the
Discriminant D = b^2 - 4ac
If D < 0, there are NO real roots
If D = 0, there is 1 real root
If D >0, there are 2 real roots

From x^2+16x+64=0 , we see that a = 1, b = 16 and c = 64
Therefore D = 16^2 -4*64 = 0
Hence there is exactly 1 real root.

How many real roots does the equation x^2y+16xy+64y=0 have if y < 0?

A. 0
B. 1
C. 2
D. 3
E. Infinite

Hi Vincen,
Let's take a look at your question.

$$x^2y+16xy+64y=0$$
$$y\left(x^2+16x+64\right)=0$$
$$x^2+16x+64=0$$

To find the number of roots of a quadratic equation, we use the discriminant.

If Discriminant < 0, there are two distinct complex roots of the quadratic equation.
If Discriminant = 0, there is only one real root of the quadratic equation.
If Discriminant > 0, there are two distinct real roots of the quadratic equation.

So let's find the discriminant of the given equation.
$$Discriminant=b^2-4ac$$
$$=\left(16\right)^2-4\left(1\right)\left(64\right)=256-256=0$$

Since, discriminant is zero, therefore, the given equation has only one root.
Therefore, Option B is correct.

Hope it helps.
I am available if you'd like any follow up.