# Thread: Tangent to a curve

1. ## Tangent to a curve

For curve $x^3 + y^3 - 9xy = 0$ find x co-ordinates where the tangent to the curve is horizontal.

I differentiated: $\frac{d}{dx}[x^3 + y^3 - 9xy] = \frac{3y - x^2}{y^2 - 3x}$

Let this equal zero for horizontal tangents, but I don't know how to solve from here, the answer is $54^{\frac{1}{3}}$ but my answers always seem to have a y term in them.

$3y - x^2 = 0$
$y = \frac{x^2}{3}$
$x^3 + \left(\frac{x^2}{3}\right)^3 - 9x\left(\frac{x^2}{3}\right) = 0$