If x + y + z = 0, then what is the value of $$\frac{x^{2}}{yz}+\frac{y^{2}}{xz}+\frac{z^{2}}{xy}$$?

Given x + y + z = 0. So, x + y = -z, y + z = -x, z + x = -y -----------(1)

$$\frac{x^{2}}{yz}+\frac{y^{2}}{xz}+\frac{z^{2}}{xy}$$ 

$$\frac{x^{3} + y^{3} + z^{3}}{xyz}$$ -----------(2)

We know that, $$a^{3} + b^{3} + c^{3} = (a+b+c)^{3} - 3ab(a+b) - 3bc(b+c) - 3ac(a+c) - 6abc$$

Hence, equation (2) can be written as,

$$\frac{(x+y+z)^{3} - 3xy(x+y) - 3yz(y+z) - 3zx(x+z) - 6xyz}{xyz}$$

Now substitute equation (1) in the above equation,

$$\frac{(0)^{3} + 3xy(z) + 3yz(x) + 3zx(y) - 6xyz}{xyz}$$

$$\frac{3xyz}{xyz}$$ = 3

Hence, option D is the correct answer.