Given function is f(x,y)=x2+y2xy3 when (x,y)=(0,0).
For this case, denominator is not zero and also enumerator and denominator are continuous, so given function is differential for this case.
So, fx(x,y)=(x2+y2)2(x2+y2)(y3)−(xy3)(2x)=(x2+y2)2y3(y2−x2).
Given f(x,y)=0 for (x,y)=(0,0)
Now fx(0,0)=limh→0hf(0+h,0)−f(0,0)=limh→0h0=0 .
Also, lim(x,y)→(0,0)fx(x,y)=0 since numerator power is more than enumerator.
So, fx(x,y) is continuous at (0,0).
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