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