<s>[INST] To show that a partial dertivative (of a piecewise function) is continuous at $0$ 
$$f(z)=\cases{\frac{x^4-6x^2y^2+y^4}{x^2+y^2}
 +i\frac{4xy(x^2-y^2)}{x^2+y^2},& $z\ne0$\cr 0, &$z=0$}$$
Let $u=\Re(f)$.

I have shown from first principles that $\frac{\partial u}{\partial x} = 0$ at $0$. To go on to show that   $\frac{\partial u}{\partial x}$ is continuous at $0$, I have to show that $\frac{\partial u}{\partial x}\to 0$ near $0$, right? But while the numerator of this partial clearly approaches $0$ near $0$, so does the denominator...

It would be most helpful if someone can demonstrate how to prove that
  $\frac{\partial u}{\partial x}$ is continuous at $0$.

P.S. Will someone also help to link this post to my earlier one: Show that this piecewise function is differentiable at $0$
 [/INST] $$\frac{\partial u}{\partial x}=\frac{(x^2+y^2)(4x^3-12xy^2)-2x(x^4-6x^2y^2+y^4)}{(x^2+y^2)^2}=$$
$$=\frac{2x^5-14xy^4+4x^3y^2}{(x^2+y^2)^2}=2x\frac{x^4+2x^2y^2-7y^4}{(x^2+y^2)^2}\implies$$
$$\left|\;\frac{\partial u}{\partial x}\;\right|\le|2x|\frac{x^4+2x^2y^2+7y^4}{(x^2+y^2)^2}=|2x|\left(1+6\frac{y^4}{(x^2+y^2)^2}\right)\xrightarrow[(x,y)\to(0,0)]{}0$$
Hint for the last part: what's inside the parentheses is bounded. Use polar coordinates, for example.
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