We are given the equation:

$x^2 + 3y^2 = 3$

We need to find $y''$ (the second derivative of $y$ with respect to $x$) using implicit differentiation. Here are the steps:

Step 1: Differentiate both sides of the equation implicitly with respect to $x$.

$\frac{d}{dx} (x^2 + 3y^2) = \frac{d}{dx}(3)$

Using the chain rule for the $y^2$ term:

$\frac{d}{dx}(x^2) + \frac{d}{dx}(3y^2) = 0$

$2x + 3 \cdot 2y \cdot \frac{dy}{dx} = 0$

Simplifying:

$2x + 6y \frac{dy}{dx} = 0$

Step 2: Solve for $\frac{dy}{dx}$.

$6y \frac{dy}{dx} = -2x$

[ \frac{dy}{dx} = \frac{-2x}{6y} =