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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?

Al-amin Chowdhury
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    I would suggest that you show how you got your values for the derivatives. I don't get infinities at all. Also, sorry, but something is clearly wrong here because $dy/dx$ and $dx/dy$ cannot be infinite at the same time. – bob.sacamento Jun 15 '21 at 17:00
  • @bob.sacamento I have edited the question and the solving process is still unknown to me. – Al-amin Chowdhury Jun 16 '21 at 08:56
  • OK. That's better. Zero divided by zero is absolutely not always infinity! You are going to have to find the limit of the derivative as $x$ approaches zero. Others might have a better way, but I would suggest starting with your original function and solving for $y$ in terms of $x$. It will be a bit of a mess, but it can be done since it will just boil down to a quadratic equation for $y$. Then see what happens to the derivative of that function as $x$ approaches zero. – bob.sacamento Jun 16 '21 at 14:53

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