Aloha :)
Aus den partiellen Ableitungen$$f(x,y)=\frac{x-y}{x+y}\quad\Rightarrow\quad f(2,2)=0$$$$f_x(x,y)=\frac{2y}{(x+y)^2}\quad\Rightarrow\quad f_x(2,2)=\frac{1}{4}$$$$f_y(x,y)=-\frac{2x}{(x+y)^2}\quad\Rightarrow\quad f_y(2,2)=-\frac{1}{4}$$$$f_{xx}(x,y)=-\frac{4y}{(x+y)^3}\quad\Rightarrow\quad f_{xx}(2,2)=-\frac{1}{8}$$$$f_{yx}(x,y)=\frac{2(x-y)}{(x+y)^3}\quad\Rightarrow\quad f_{yx}(2,2)=0$$$$f_{yy}(x,y)=\frac{4x}{(x+y)^3}\quad\Rightarrow\quad f_{yy}(2,2)=\frac{1}{8}$$erhalten wir die gesuchte Taylor-Entwicklung
$$f(x,y)=f(2,2)+f_x(2,2)\cdot(x-2)+f_y(2,2)\cdot(y-2)$$$$\quad+\frac{1}{2}f_{xx}(2,2)\cdot (x-2)^2+f_{yx}(2,2)\cdot(x-2)(y-2)+\frac{1}{2}f_{yy}(2,2)\cdot(y-2)^2$$$$\quad=\frac{1}{4}\cdot(x-2)-\frac{1}{4}\cdot(y-2)-\frac{1}{16}\cdot(x-2)^2+\frac{1}{16}\cdot(y-2)^2$$$$\quad=\frac{x-y}{2}+\frac{y^2-x^2}{16}$$
