Aloha :)
$$f(x;y)=\ln\left(\frac{x^2}{e^{2y}}\right)-\frac xy=\ln\left(x^2\,e^{-2y}\right)-\frac xy$$
$$\frac{\partial f}{\partial x}=\underbrace{\frac{1}{\frac{x^2}{e^{2y}}}}_{\text{äußere}}\cdot\underbrace{\frac{2x}{e^{2y}}}_{\text{innere}}-\frac1y=\frac{e^{2y}}{x^2}\cdot\frac{2x}{e^{2y}}-\frac1y=\frac2x-\frac1y$$
$$\frac{\partial f}{\partial y}=\underbrace{\frac{1}{\frac{x^2}{e^{2y}}}}_{\text{äußere}}\cdot\underbrace{x^2\,e^{-2y}\cdot(-2)}_{\text{innere}}+\frac{x}{y^2}=\frac{e^{2y}}{x^2}\cdot\frac{x^2}{e^{2y}}\cdot(-2)+\frac{x}{y^2}=\frac{x}{y^2}-2$$