Derivative for $x^3+ax^2y+bxy^2+y^3=0$ Let $x^3+ax^2y+bxy^2+y^3=0$. Find $y'$ in point $M(0;0)$
My work:
$$3x^2+2axy+ax^2y'+by^2+2bxyy'+3y^2y'=0$$
$$y'=-\frac{3x^2+2axy+by^2}{ax^2+2bxy+3y^2}$$
How find $y'(0;0)$?
 A: You can try plugging in $x=0,y=0$ to your equation for $y'$.  That doesn't work, it has form $0/0$.  You can try l'Hopital's rule on that; looks complicated.  
Here is another way.
At the point $(0,0)$, the tangent $y'$ will be the limit of $y/x$.  So write
$$
x^3+ax^2y+bxy^2+y^3=0
\\
1+a(y/x)+b(y/x)^2+(y/x)^3=0
$$
and in the limit,
$$
1+ay'+b(y')+(y')^3 = 0
$$
a cubic equation to solve for the vaue of $y'$ at $(0,0)$.  
There could be three solutions to the cubic equation.  So the curve could pass through $(0,0)$ three times with different slopes.  
In fact, in general $x^3+ax^2y+bxy^2+y^3=0$ is three straight lines through the origin.  You would need lower-degree terms to get a more interesting curve.
A: Hint. Let $y=mx$ be the equation of the tangent line at $(0,0)$ where $m=y'(0)$, then $m$ satisfies cubic equation
$$1+am+ bm^2+m^3=0.$$
P.S. Note that the number of tangent line  through the origin can be $1$, $2$ or $3$ (it depends on the number of real roots of the cubic). For example if $a=0$ and $b=-2$ then
$$x^3+ax^2y+bxy^2+y^3=(y-x)(y-\frac{1+\sqrt{5}}{2}\, x)(y-\frac{1-\sqrt{5}}{2}\, x)=0$$
and we have three tangents at $(0,0)$ (actually the curve is the union of those three lines)
$$y=x,\quad y=\frac{1+\sqrt{5}}{2}\, x,\quad y=\frac{1-\sqrt{5}}{2}\, x.$$
