My aim is to show that the function $$f(x+iy)=x^3+ax^2y+bxy^2+cy^3$$ is analytic only if $a=3i$ , $b=-3$ , $c=-i$.
For these values of $a$, $b$ and $c$, that $f$ is analytic follows from Cauchy-Riemann equation.
But I unable to test the only if condition.
Just use the Cauchy-Riemann equations again. They are a necessary condition, and they give you a system of equations which you can solve for $a$, $b$, and $c$.
Say $f(x+iy)=u(x,y)+iv(x,y)$ then $u(x,y)=x^3+a_1x^2 y+b_1xy^2+c_1y^3$ and $v(x,y)=a_2x^2y+b_2xy^2+c_2y^3$ where $a=a_1+ia_2,b=b_1+ib_2$ and $c=c_1+ic_2$. We know that $f$ is analytic in some region if and only if all the partial derivatives of $u$ and $v$ exist and continuous and the CR-equations are satisfied at every point of the region. So apply the CR-equations to get $$3x^2+2a_1xy+b_1y^2=a_2x^2+2b_2xy+3c_2y^2$$ and $$a_1x^2+2b_1xy+3c_1y^2=-(2a_2xy+b_2y^2)$$ for all $x,y\in \Bbb R$ as $f$ is analytic. Then you can equate the coefficients to the polynomials in both sides of each of these equations and can conclude that $a_2=3,2b_2=2a_1,b_1=3c_2, a_1=0,2b_1=-2a_2$ and $3c_1=-b_2$. From this it follows that $a_1=0=b_2,b_1=-a_2=-3,c_2=-\frac{1}{3}b_1=-1,c_1=-\frac{1}{3}b_2=0.$ So $a=3i,b=-3$ and $c=-i$.
Here is another way:
Suppose $f$ is analytic. Note that $|f(z)| \le L(1+|z|^3)$ for some $L>0$, then it follows from Cauchy's estimates that $f$ is a polynomial of degree at most 3, that is $f(z) = \sum_{k=0}^3 a_k z^k$.
Evaluating along the real line (that is, $y=0$) gives $f(x) = \sum_{k=0}^3 a_k x^k$, and we know that $f(x) = x^3$, hence we have $a_3=1$ and $a_k = 0$ for $k = 0,1,2$. That is, we have $f(z) = z^3$.
Now expand $(x+iy)^2$, then comparing coefficients shows that $a,b,c$ must have the values above.
Note: The above uses the fact that the functions $(x,y) \mapsto x^my^n$ are linearly independent.
