I have doubts in the question, so I have given two answers.
If the given function is
f(x)={2x+2yxy0if(x,y)=(0,0)if(x,y)=(0,0)
Consider the sequence xn=(n1,n−1) in R2 .
Then limn→∞xn=(0,0). But f(xn)=n2−n2(n1)(n−1)
⟹f(xn)=0−1 .
which does not exist.
Hence f(x,y) is not continuous at (0,0).
If the function is f(x)={x2+y2xy0if(x,y)=(0,0)if(x,y)=(0,0)
Consider the sequence xn=(n1,n1) in R2.
Then limx→∞xn=(0,0). But f(xn)=n22n21=21 ,which is a constant function and converge to 21 .
∴f(xn)↛f(0) as n→∞ .
Hence f(x,y) is not continuos at (0,0).
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