<s>[INST] 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 $xy=0$
 [/INST] 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.
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