Since $$\lim_{(x,y)\to(0,0)}\frac{x^4-y^4}{x^2+y^2}=\lim_{(x,y)\to(0,0)}\frac{(x^2+y^2)(x^2-y^2)}{x^2+y^2}=\lim_{(x,y)\to(0,0)}(x^2+y^2)=0$$
and $$\lim_{(x,y)\to(0,0)}\frac{x^4}{x^2+y^2}=\lim_{(x,y)\to(0,0)}x^2\cdot\frac{x^2}{x^2+y^2}=0$$ because $$\left|\frac{x^2}{x^2+y^2}\right|\leq1$$ and $$\lim_{(x,y)\to(0,0)}x^2=0$$ I may conclude that both functions $$f(x,y)=\left\{\begin{array}{ll}\displaystyle\frac{x^4-y^4}{x^4+y^4}, & (x,y)\neq(0,0) \\ 0, &(x,y)=(0,0)\end{array}\right.$$ and $$g(x,y)=\left\{\begin{array}{ll}\displaystyle\frac{x^4}{x^4+y^4}, & (x,y)\neq(0,0) \\ 0, &(x,y)=(0,0)\end{array}\right.$$ are continuous at $(x,y)=(0,0)$.