For $f(z)=2x^3+3iy^2$, $f'(x+ix^2)=?$

Question

For $f(z)=2x^3+3iy^2$, f'(x+ix^2)=?

I said this must be analytic and provide Cauchy-Riemann equations

And my answer ==> $u_x=6x^2$, $v_y=6y$, $u_y=0$, $v_x=0$

So $6x^2=6y$

==> $x^2=y$

==>$z=x+iy=x+ix^2$ Then the question is turning to $f'(z)$

And $f'(z)$=$6x^2+12x^3i$

Is my answer true or am i doing something wrong? I am not sure.

Answer 1

Since $f=2x^3+3x^4i$, what you've calculated in the last step is $f_x$. But we want$$f_z=\frac{f_x}{z_x}=\frac{6x^2+12x^3i}{1+2xi}=6x^2.$$