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.

If we consider the differential $(6x-2)dx+8ydy$, then at $(0,0)$, we see $(6x-2) \neq 0$. By the implicit function theorem, we may write $x$ as a function of $y$. 
Complete the square:
$$3x^2 -2x +4y^2=0$$
$$3(x-\frac{1}{3})^2+4y^2=\frac{1}{3}$$
From here we get:
$$x(y)=\frac{1}{3}+\sqrt{\frac{1}{9}-\frac{4}{3}y^2}$$ 
Let $r:(-\epsilon,\epsilon) \rightarrow \mathbb{R}^{2}$ be given by 
$$r(y) =(x(y),y)$$
The curvature is given by the general formula:
$$\kappa=\frac{|x'y''-x''y'|}{(x'^{2}+y'^{2})^{\frac{3}{2}}}$$
Using our parameterization, we have:
$$\kappa(y)=\frac{|\frac{d^2x}{dy^2}|}{(1+(\frac{dx}{dy})^2)^\frac{3}{2}}$$
Now $\frac{dx}{dy}=0$ when $y=0$. If we compute $\frac{d^2
x}{dy^2}$ we have:
$$\frac{d^2
x}{dy^2}=y(junk)+\frac{4}{3}\bigg(\frac{1}{9}-(\frac{4}{3}y^2)\bigg)^\frac{-1}{2}$$
Evaluating at $y=0$ we find,
$$\kappa(0)=4$$.
Thus $\rho= \frac{1}{\kappa}=\frac{1}{4}$
