Answer
(a)
$$\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}=2x.$$
(b)
$$\lim_{\Delta y\to 0}\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}=-4.$$
Work Step by Step
We have that
$$f(x+\Delta x,y)=(x+\Delta x)^2-4y=x^2+2x\Delta x+\Delta x^2-4y,$$
and
$$f(x,y+\Delta y)=x^2-4(y+\Delta y)=x^2-4y-4\Delta y.$$
(a)
$$\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} =\\\lim_{\Delta x\to 0} \frac{x^2+2x\Delta x+\Delta x^2-4y-(x^2-4y)}{\Delta x}=\\\lim_{\Delta x\to 0} \frac{2x\Delta x+\Delta x^2}{\Delta x}=\\
\lim_{\Delta x\to 0} (2x+\Delta x) = 2x,$$
where in the last step we used method of substitution.
(b)
$$\lim_{\Delta y\to 0}\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}=\\\lim_{\Delta y\to 0}\frac{x^2-4y-4\Delta y- (x^2-4y)}{\Delta y} =\\\lim_{\Delta y\to 0}-\frac{4\Delta y}{\Delta y}=\lim_{\Delta y\to 0}-4 = -4, $$
where in the last step we used that the limit of a constant is that constant.
