Find the radius of curvature at the origin of the curve $$y^2-2xy-3x^2-4x^3-x^2y^2 = 0$$
In this math,
$$\frac {dy}{dx} = \frac {y+3x+6x^2+xy^2}{y-x-yx^2}$$ and $$\frac {dx}{dy} = \frac {y-x-yx^2}{y+3x+6x^2+xy^2}$$
At $(0, 0)$ both $$ \frac {dy}{dx} = \frac {0}{0} $$ and $$ \frac {dx}{dy} = \frac {0}{0}$$
So, I am not able to apply the equations $$\rho=\dfrac{{(1+y_1^2)}^{\frac{3}{2}}}{y_2}$$ and
$$\rho=\dfrac{{(1+x_1^2)}^{\frac{3}{2}}}{x_2}$$
Then, how do I find the tangent and radius of the curvature?