Answer
(a)
$$\lim_{\Delta x\to 0}\frac{f(x+\Delta x,y)-f(x,y)}{\Delta x}=-\frac{1}{(x+y)^2}.$$
(b)
$$\lim_{\Delta y\to 0}\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}=-\frac{1}{(x+y)^2}.$$
Work Step by Step
We have that
$$f(x+\Delta x,y)=\frac{1}{x+\Delta x+y}.$$
and
$$f(x,y+\Delta y)=\frac{1}{x+y+\Delta y}.$$
(a)
$$\lim_{\Delta x\to 0}\frac{f(x+\Delta x,y)-f(x,y)}{\Delta x} =\lim_{\Delta x\to 0}\frac{\frac{1}{x+\Delta x+y}-\frac{1}{x+y}}{\Delta x}= \lim_{\Delta x\to 0}\frac{\frac{x+y-x-\Delta x-y}{(x+y)(x+\Delta x +y)}}{\Delta x}=\\\lim_{\Delta x\to 0}\frac{-\Delta x}{\Delta x(x+y)(x+\Delta x+y)} = \lim_{\Delta x\to 0}\frac{-1}{(x+y)(x+\Delta x +y)} = -\frac{1}{(x+y)(x+0+y)} = -\frac{1}{(x+y)^2}.$$
where in the last two steps we used the method of substitution and then calculated the limit.
(b)
$$\lim_{\Delta y\to 0}\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}= \lim_{\Delta y\to 0}\frac{\frac{1}{x+y+\Delta y}-\frac{1}{x+y}}{\Delta y}= \lim_{\Delta y\to 0}\frac{\frac{x+y-x-y-\Delta y}{(x+y)(x+y+\Delta y)}}{\Delta y}=\\\lim_{\Delta y\to 0}\frac{-\Delta y}{\Delta y(x+y)(x+y+\Delta y)} = \lim_{\Delta y\to 0}\frac{-1}{(x+y)(x+y+\Delta y)} = -\frac{1}{(x+y)(x+y+0)} = -\frac{1}{(x+y)^2}.$$
where in the last two steps we used the method of substitution and then calculated the limit.
