y=x2-3x=x(x-3), y2=x2(x-3)2; x2+y2=9, y2=9-x2=(3-x)(3+x).
(3-x)(3+x)=x2(x-3)2. x=3 is clearly one solution, making y=0.
Dividing by x-3:
-(3+x)=x3-3x2,
x3-3x2+x+3=0.
To solve the cubic we can use Newton's Method: xn+1=xn-(xn3-3xn2+xn+3)/(3xn2-6xn+1) where the denominator is the derivative of the numerator wrt x.
This is an iterative process which, given a starting value x0, will quickly yield a solution for x. Let x0=0.
x1=-3, x2=-42/23, x3=-1.1467..., x4=-0.8423..., x5=-0.7728..., x6=-0.7693..., x7=-0.76929..., x8=-0.7692923642 (stable solution).
From this value of x we get y=2.899687789. So the solution is: (x,y)=(3,0) or (-0.7693,2.8997) approx.