Finding $y^{\prime \prime}$ of $2x^2+3y^2=4$ 
Find $y^{\prime}$ and $y^{\prime \prime}$ of $2x^2+3y^2=4$  

$$y^{\prime}=\dfrac{d}{dx}(2x^2)+\dfrac{d}{dx}(3y^2)=\dfrac{d}{dx}(4)$$
$$4x+6yy^{\prime}=0$$
$$y^{\prime}=\dfrac{-2x}{3y}$$ 
This is how I started finding $y^{\prime\prime}$:
$$y^{\prime\prime}=\dfrac{3y \dfrac{d}{dx}(-2x) - (-2x)\dfrac{d}{dx}(3y)}{(3y^2)}$$
$$\dfrac{3y(-2)-[-2x(3y^{\prime})]}{9y^2}$$
$$\dfrac{-6y-[-6xy^{\prime}]}{9y^2}$$
This isn't right since the correct answer is $\dfrac{-6y^2+4x^2}{9y^3}$
Can you please show how to find $y^{\prime\prime}?$ Thank you. 

Note that 
$$6x \cdot \frac{-2x}{3y} = \frac{-12x^{2}}{3y} = \frac{-4x^{2}}{y}$$
Therefore, we have that
$$y'' = \frac{-6y + \frac{-4x^{2}}{y}}{9y^{2}}$$
Multiplying by $\frac{y}{y} = 1$, we find:
$$y'' = \frac{-6y^{2} - \frac{4x^{2}}{y}\cdot y}{9y^{3}} = \frac{-6y^{2}-4x^{2}}{9y^{3}}$$
