Find the equation of tangents for $x^3+y^3-3xy=0$ at $x=0,y=0$ $$x^3+y^3-3xy=0$$
Find the equation of tangents at $x=0,y=0$
My attempt is as follows:-
Attempt $1$:
$$3x^2+3y^2\dfrac{dy}{dx}-3\left(x\dfrac{dy}{dx}+y\right)=0$$
$$\dfrac{dy}{dx}(y^2-x)=y-x^2$$
$$\dfrac{dy}{dx}=\dfrac{y-x^2}{y^2-x}$$
but when placing $x=0,y=0$ we are getting undefined quantity.
But actual answer is $xy=0$, how can we proceed here?
Attempt $2$: (Parametric method)
$$\dfrac{y}{x}=t$$
$$y=xt$$
Putting this in the original equation
$$x^3+x^3t^3-3x^2t=0$$
$$x^2(x+xt-t)=0$$
$$x=\dfrac{t}{t+1}$$
$$y=\dfrac{t^2}{t+1}$$
$$\dfrac{dy}{dx}=\dfrac{\dfrac{2t(t+1)-t^2}{(t+1)^2}}{\dfrac{t+1-t}{(t+1)^2}}$$
$$\dfrac{dy}{dx}=t^2+2t$$
$x=0$, then $\dfrac{t}{t+1}=0 \implies t=0$
$y=0$, then $\dfrac{t^2}{t+1}=0$ also $\implies t=0$
So we are getting slope as $0$, hence $y=0$ can be the answer, but actual answer is $xy=0$
 A: Continue with what you obtained 
$$y'= \frac{y-x^2}{y^2-x} $$
and note that the near the origin, $y' = \frac yx$, or 
$$\frac{y-x^2}{y^2-x}  = \frac yx \implies x^3+y^3 = 2xy$$
Then, substitute $x^3+y^3 - 3xy=0$ to obtain the equation near the origin, 
$$xy=0$$
which represents the $x$- and the $y$-axes and also the tangent lines of the given curve at the origin.
A: The issue is, as you have noted, the tangent line is ill-defined at $(0,0)$. See the plot at this link. There are two distinct tangent lines at $(0,0)$: $x=0$ and $y=0$ (hence the 'answer' of $xy=0$), depending on where you are 'in time', i.e. imagine tracing the curve out in time, at one time (let's say drawing left to right), $y=0$ would be the appropriate tangent while at another $x=0$ would be the appropriate tangent. If this curve were given parametrically, we would have an 'obvious' choice but as it stands there is no way of choosing one tangent over the other. This is the same issue one runs into in definition the addition law for elliptic curves in the case of the nodal cubic. As a final note, a parametrization is given at this link, where you can then as you have derive the appropriate tangents. 
