Find the radius of curvature at the origin 
Find the radius of curvature at the origin for the curve $3x^2+4y^2=2x$.

Below is my attempt:
We know that the radius of curvature $\rho=\dfrac{{(1+y_1^2)}^{\frac{3}{2}}}{y_2}$ .
Computing $y_1$ at $(0,0)$ : Differentiating we get $3x+4y\frac{dy}{dx}=1\implies \dfrac{dy}{dx}=\dfrac{1-3x}{4y}.$
How can I evaluate $\dfrac{dy}{dx}$ at the point $(0,0)$?
Please help.

Since, $$\frac{dy}{dx}$$is not defined here, i.e., y axis is tangent to the curve.
No one above is talking about Newtonian method, so I am going with that - it says if curve passes through origin and y axis(or x axis) is tangent to the curve, then radius of curvature at origin = $$lim(x \rightarrow 0) \frac{y²}{2x}$$
We have, $$3{x}^2 - 2x + 4{y}^2 = 0 $$
Dividing it by 2x,
$$\frac{3x}{2} - 1 + 4\frac{y^2}{2x} =0$$
When x --> 0,
$$ -1 + 4\rho = 0$$
$$ \rho = \frac{1}{4}$$
