Where is $|z|^4 + z^2$ differentiable? Where is $|z|^4 + z^2$ differentiable?

well, I guess i suceeded at proving that its differentiable at $z_0=0$.
im stuck with the case where $z_o\neq0$. how do I do this by definition? (without using cauchy reimann's equations).
here is what I statrted and couldn't finish: 


Edit: lets say im ignoring $z^2$ because its a ploynomial. here is what I get: 

and yet, I still need to show that this limit doesn't exist.I can't see how.
$f(x,y)$is  differentiable if $$f(x,y)=f(x+iy,0)$$
$$f(x,y)=(\sqrt{x^2+y^2})^4+(x+iy)^2=(x^2+y^2)^2+x^2-y^2+i(2xy)$$so check WHEN $f(x,y)=f(x+iy,0)$
$$f(x+iy,0)=((x+iy)^2+0^2)^2+(x+iy)^2-0^2+i(2x0)=\\
(x^2-y^2+i(2xy))^2+x^2-y^2+i(2xy)$$
now $$(x^2+y^2)^2+x^2-y^2+i(2xy)=((x+iy)^2+0^2)^2+(x+iy)^2-0^2+i(2x0)\\
x^4+y^4+2x^2y^2+x^2-y^2+i(2xy)=x^4+4x^3(iy)+6x^2(-y^2)+4x(-iy^3)+y^4+x^2-y^2+i(2xy)$$simplify
$$2x^2y^2=4x^3(iy)-6x^2y^2-4x(iy^3)\\8x^2y^2=4xy(ix^2-iy^2)$$
so it is differentiable when 
$$\begin{cases}xy=0 &amp; x=0 &amp;y=0\\xy\neq0 &amp; 2xy=i(x^2-y^2)&amp; real \neq image\end{cases}$$so $$f(x,y)$$is differentiabale when $xy=0$
As suggested in the comments, it suffices we consider $|z|^4$. Write $f(z) = |z|^4 = \bar z^2z^2$. Differentiating with respect to $\bar z$ gives $\partial_{\bar z}f(z) = 2\bar z z^2$ and this vanishes only when $z=0$. So $f$ is differentiable only at the origin.