To find the tangent, differentiate the given curve implicitly:
`rArr y'=(y-3x^2)/( 3y^2-x)`
This is the gradient of tangent to the given curve.
By the condition of the problem, the tangent is parallel to the x-axis. So its gradient must be zero.
So, `(y-3x^2)/( 3y^2-x)=0`
Putting this value of y in the curve,
This equation has two solutions: `x=0` , and
When `x=0, y=3*0^2=0` , and
when `x= root(3)(2)/3, y= 3*(root(3)(2)/3)^2=root(3)(4)/3` .
Thus, there will be two points in the first quadrant where our tangent will be parallel to the curve `x^3-xy+y^3=0` , the coordinates of which are `(0,0)` and `(root(3)(2)/3, root(3)(4)/3)`.