Let's find a limit of this function at (0,0)
limx,y→0xy3x2+y2=[y=xc;c∈R]=limx,y→0x3c+1x2(1+x2(c−1)=0=f(0,0)\lim\limits_{x,y\to 0}\frac{xy^3}{x^2+y^2}=[y=x^c;c\in R]=\lim\limits_{x,y\to 0}\frac{x^{3c+1}}{x^2(1+x^{2(c-1})}=0=f(0,0)x,y→0limx2+y2xy3=[y=xc;c∈R]=x,y→0limx2(1+x2(c−1)x3c+1=0=f(0,0)
Therefore the function is continuous at (0,0)
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