Find the equation of tangent at origin to the curve $y^2=x^2(1+x+x^2)$ How do I find the equation of tangent at $(0,0)$ to the curve $y^2=x^2(1+x+x^2)$ ?
Differentiating and putting the value of $x$ and $y$ gives an indeterminate form. 
Can we trace the curve and geometrically make tangents and find their equation ?

$$y^2=x^2(1+x+x^2)\implies 2yy'=2x(1+x+x^2)+x^2(1+2x)=4x^3+3x^2+2x\implies$$
$$y'(0)=\lim_{x\to 0}\frac{4x^3+3x^2+2x}{2y}=\pm\lim_{x\to0}\frac{4x^2+3x+2}{2\sqrt{1+x+x^2}}=\pm1$$
and thus the tangent doesn't exist since $\;y=|x|\sqrt{1+x+x^2}\;$ and $\;\frac x{|x|}=\pm1\;$
