Answer
(a)
$$\lim_{\Delta x\to0} \frac{f(x+\Delta x,y)-f(x,y)}{\Delta x}=3+y.$$
(b)
$$\lim_{\Delta y\to 0}\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y} = x-2.$$
Work Step by Step
We have
$$f(x+\Delta x,y)=3(x+\Delta x)+(x+\Delta x)y-2y=3x+3\Delta x+xy+\Delta xy-2y=3x+xy-2y+3\Delta x+\Delta xy.$$
and
$$f(x,y+\Delta y)=3x+x(y+\Delta y)-2(y+\Delta y)=3x+xy-2y+x\Delta y-2\Delta y.$$
(a)
$$\lim_{\Delta x\to0} \frac{f(x+\Delta x,y)-f(x,y)}{\Delta x} = \lim_{\Delta x\to0}\frac{3x+xy-2y+3\Delta x+\Delta xy-(3x+xy-2y)}{\Delta x} = \lim_{\Delta x\to0}\frac{3\Delta x+\Delta xy}{\Delta x} = \lim_{\Delta x\to0} (3+y) = 3+y,$$
where in the last step we used that the limit of the constant is that constant itself (any expression containing only $y$ is treated as a constant.)
(b)
$$\lim_{\Delta y\to 0}\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}=\lim_{\Delta y\to 0}\frac{3x+xy-2y+x\Delta y-2\Delta y-(3x+xy-2y)}{\Delta y} = \lim_{\Delta y\to 0}\frac{x\Delta y-2\Delta y}{\Delta y} = \lim_{\Delta y\to 0}(x-2) = x-2$$
where in the last step we used that the limit of the constant is that constant itself (any expression containing only $x$ is treated as a constant.)
