(a)Here is discontinuous function at (0,0) having partial derivatives at (0,0)
f(x,y)={x2+y22xy0(x,y)=(0,0)(x,y)=(0,0)
showing f is not continuous at (0,0)
choosing the path x = 0 we see that f(0,y)=0, so limy→0f(0,y)=0.
choosing the path x=y we see that f(x,x)=2x22x2=1, so limx→0f(x,x)=1. The two path theorem implies that lim(x,y)→(0,0)f(x,y) does not exist
(b) Finding fx(0,0) and fy(0,0)
the partial derivatives are defined at (0,0)
fx(0,0)=limh→0h1[f(0+h,0)−f(0,0)]=limh→0h1[0−0]=0
fy(0,0)=limh→0h1[f(0,0+h)−f(0,0)]=limh→0h1[0−0]=0
Therefore fx(0,0)=fy(0,0)=0
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