Find \( y^{\prime \prime} \) by implicit differentiation. \[ 4 x^{3}-5 y^{3}=2 \] \( y^{\prime \prime}=\square \)

To find \( y'' \) by implicit differentiation, we start with the equation: \[ 4x^3 - 5y^3 = 2. \] First, we differentiate both sides with respect to \( x \): \[ \frac{d}{dx}(4x^3) - \frac{d}{dx}(5y^3) = \frac{d}{dx}(2). \] This yields: \[ 12x^2 - 15y^2 \frac{dy}{dx} = 0. \] Solving for \( \frac{dy}{dx} \): \[ 15y^2 \frac{dy}{dx} = 12x^2 \] \[ \frac{dy}{dx} = \frac{12x^2}{15y^2} = \frac{4x^2}{5y^2}. \] Now we differentiate \( \frac{dy}{dx} \) again to find \( y'' \): Using the quotient rule: \[ y'' = \frac{d}{dx}\left(\frac{4x^2}{5y^2}\right). \] Applying the quotient rule gives: \[ y'' = \frac{5y^2 \cdot \frac{d}{dx}(4x^2) - 4x^2 \cdot \frac{d}{dx}(5y^2)}{(5y^2)^2}. \] Calculating \( \frac{d}{dx}(4x^2) = 8x \) and \( \frac{d}{dx}(5y^2) = 10y \frac{dy}{dx} \), we have: \[ y'' = \frac{5y^2 \cdot 8x - 4x^2 \cdot 10y \frac{dy}{dx}}{25y^4}. \] Substituting \( \frac{dy}{dx} = \frac{4x^2}{5y^2} \): \[ y'' = \frac{40xy^2 - 40x^2 \cdot \frac{4x^2}{5y^2}}{25y^4}. \] Simplifying this, gives: \[ y'' = \frac{40xy^2 - \frac{160x^4}{5y^2}}{25y^4}, \] which simplifies further to: \[ y'' = \frac{200xy^4 - 160x^4}{125y^6}. \] Thus, the final expression for \( y'' \) is: \[ y'' = \frac{200xy^4 - 160x^4}{125y^6}. \]