Calculate the second derivative of $5x^3+9y^3=19$.

$\frac{d^{2}y}{d{x}^2}=\frac{50x^4+90x{y}^3}{81y^5}$

$\frac{d^{2}y}{d{x}^2}=-\frac{50x^4+90x{y}^3}{81y^5}$

$\frac{d^{2}y}{d{x}^2}=\frac{50x^4-90x{y}^3}{81y^5}$

$\frac{d^{2}y}{d{x}^2}=\frac{81y^5}{50x^4+90x{y}^3}$