If at any particular point the function does not satisfy the Cauchy-Riemann equations, then its complex derivative cannot exist at that point. So we check the Cauchy Riemann equations and see that $$u_x = 3x^{2}+y^{2}~~,~~u_y = 2xy~~,~~v_x = 2xy~~,~~ v_y = x^{2}+3y^{2}$$
Equipping we see that $$3x^{2}+y^{2} = x^{2}+3y^{2} ~\text{and}~ 2xy = -2xy$$
However my problem came when i tried to solve the equation above, i somehow get that the Cauchy Riemann equations are SATISFIED at the points $(0,0)$, does that suggest that function $z^{2}\overline{z}$ actually has a derivative at $z_0 = 0$? I am really confused, please help.
P.S Maybe some kind souls can show their working for me?