Aloha :)
In beiden Fällen die erste Ableitung mit der Kettenregel \((\,\ln(y(x)\,)'=\frac{y'(x)}{y(x)}\) und die zweite Ableitung mit der Quotientenregel \((\frac{u}{v})'=\frac{u'v-uv'}{v^2}\).
$$\frac{\partial^2}{\partial x^2}\ln\left(\frac{x^2}{y}+3\right)=\frac{\partial}{\partial x}\left(\frac{\frac{2x}{y}}{\frac{x^2}{y}+3}\right)=\frac{\partial}{\partial x}\left(\frac{2x}{x^2+3y}\right)=\frac{2(x^2+3y)-2x\cdot2x}{(x^2+3y)^2}$$$$\phantom{\frac{\partial^2}{\partial x^2}\ln\left(\frac{x^2}{y}+3\right)}=\frac{-2x^2+3y}{(x^2+3y)^2}$$
$$\frac{\partial^2}{\partial y^2}\ln\left(\frac{x^2}{y}+3\right)=\frac{\partial}{\partial y}\left(\frac{-\frac{x^2}{y^2}}{\frac{x^2}{y}+3}\right)=\frac{\partial}{\partial y}\left(\frac{-x^2}{x^2y+3y^2}\right)=\frac{x^2(x^2+6y)}{(x^2y+3y^2)^2}$$$$\phantom{\frac{\partial^2}{\partial y^2}\ln\left(\frac{x^2}{y}+3\right)}=\frac{x^2(x^2+6y)}{y^2(x^2+3y)^2}$$
