Let $f : \Bbb C \longrightarrow \Bbb C$ (the set of all complex numbers) be defined by $$f(x+iy) = x^3 + 3xy^2 + i\ \left (y^3 + 3x^2 y \right ),\ \ i = \sqrt {-1}.$$ Let $f'(z)$ denote the derivative of $f$ with respect to $z.$
Then which one of the following statements is TRUE$?$
$(\text {A} )$ $f'(1+i)$ exists and $\left | f'(1+i) \right | = 3 \sqrt 5.$
$(\text {B})$ $f$ is analytic at the origin.
$(\text {C})$ $f$ is not differentiable at $i.$
$(\text {D})$ $f$ is differentiable at $1.$
My attempt $:$ Let $u(x,y) = x^3 + 3xy^2$ and $v(x,y) = y^3 + 3x^2y.$ Then I find that $$\begin{align*} \frac {\partial u} {\partial x} & = 3(x^2+y^2). \\ \frac {\partial u} {\partial y} & = 6xy. \\ \frac {\partial v} {\partial x} & = 6xy. \\ \frac {\partial v} {\partial y} & = 3(x^2+y^2).\end{align*}$$
So $f$ will satisfy Cauchy-Riemann equations iff $xy=0$ i.e. either $x=0$ or $y=0.$ So $f$ cannot be differentiable anywhere in the complex plane except the points on the real and imaginary axes. Since $1+i$ is lying outside the real and imaginary axes $f$ can't be differentiable at $1+i.$ So $(\text {A})$ is false. Secondly if $f$ was analytic at the origin then it has a power series expansion in some open ball centered at the origin. Hence $f$ has to be differentiable in some open ball centered at the origin. But whatever small open ball we take centered at the origin it will contain infinitely many points lying outside the real and imaginary axes where $f$ fails to be differentiable. Hence $f$ can't be analytic at the origin. So $(\text {B})$ is false. Now at the points $i$ and $1$ Cauchy-Riemann equations are satisfied. Though we can't conclude that $f$ is differentiable at those points. We have to formally check that. Now $f$ is differentiable at $i$ iff both $u$ and $v$ are differentiable at $(0,1).$ To check that we need to show that $$\begin{align*} \lim\limits_{\substack {h \to 0 \\ k \to 0}} \frac {u(h,1+k) - u(0,1) - h \frac {\partial u} {\partial x} (0,1) - k \frac {\partial u} {\partial y} (0,1)} {\sqrt {h^2 + k^2}} & = 0. \\ \lim\limits_{\substack {h \to 0 \\ k \to 0}} \frac {v(h,1+k) - v(0,1) - h \frac {\partial v} {\partial x} (0,1) - k \frac {\partial v} {\partial y}(0,1)} {\sqrt {h^2 + k^2}} & = 0. \end{align*}$$
Similarly to show that $f$ is differentiable at $1$ we need to show that $u$ and $v$ are differentiable at $(1,0)$ i.e. we need to show that $$\begin{align*} \lim\limits_{\substack {h \to 0 \\ k \to 0}} \frac {u(1+h,k) - u(1,0) - h \frac {\partial u} {\partial x} (1,0) - k \frac {\partial u} {\partial y} (1,0)} {\sqrt {h^2 + k^2}} & = 0. \\ \lim\limits_{\substack {h \to 0 \\ k \to 0}} \frac {v(1+h,k) - v(1,0) - h \frac {\partial v} {\partial x} (1,0) - k \frac {\partial v} {\partial y}(1,0)} {\sqrt {h^2 + k^2}} & = 0. \end{align*}$$
Am I going in the right direction? Is there any easier way? Any suggeston regarding this will be highly appreciated.
Thank you very much for your valuable time.
EDIT $:$ Fortunately all the double limits exist and equal to $0$ what I have just checked. Hence $f$ is differentiable at both the points $1$ and $i.$ Hence $(\text {C} )$ is false. So $(\text {D})$ is the only correct option.
First $f=(u,v)$, $u,v$ are polynomials so $f$ is Real-differentiable. Now, $u, v$ are Real-differentiable and the points where they satisfy CR equations are the real and imaginary axis. So, it is complex-differentiable at those points. It is not analytic at origin because there is no neighborhood where at all points it is complex differentiable. In fact, it is nowhere analytic.