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