Determine a complex conjugate to $u(x,y)=x^3y-xy^3$ I know $\frac{\partial^2 u}{\partial x^2}=6xy$ and $\frac{\partial ^2 u}{\partial y^2} =-6xy$ and adding these together I get 0 which tells me they are harmonic functions.
To determine the harmonic conjugate, I know that $\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=3x^2y^2-y^3$ Integrating this with respect to y gives me $v=\frac{3}{2}x^2y-\frac{y^4}{4} + f(x)$
and I know that $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=-x^3+3xy^2$. Integrating this with respect to x gives $v=\frac{-x^4}{4}+\frac{3x^2 y^2}{2}+f(y)$
Where do I go from here? 
 A: I'd rather go with the second Cauchy-Riemann equation after you did the first integration (corrected down here):
$$u_y=x^3-3xy^2=-v_x=-3xy^2-f'(x)\implies f'(x)=-x^3\implies f(x)=-\frac14x^4+K$$
and etc.
A: Start with $u=x^3y-xy^3$.  Then, $u_x=3x^2y-y^3$ and $u_y=x^3-3xy^2$.
Thus, from the Cauchy-Riemann equation $u_x=v_y$ we find that 
$$u_x=v_y \implies v=\frac32 x^2y^2-\frac14 y^4 +C_1(x)$$
Next, we use this expression for $v$ in the Cauchy-Riemann equation $u_y=-v_x$ to find that
$$v_x=-u_y \implies C_1'(x) =-x^3$$ 
whereupon we see that $C_1(x) = -\frac14 x^4+C_2$.
Putting this together we have $v=\frac32 x^2y^2-\frac14 y^4 -\frac14 x^4+C_2$.
Therefore, the complex conjugate of $f$ is 

$$\bar f=(x^3y-xy^3)-i\left(\frac32 x^2y^2-\frac14 y^4 +-\frac14 x^4\right)-iC_2$$

A: Since it is a homogeneous polynomial, meaning exponent sums $3+1=1+3=4,$ this is guaranteed to be the imaginary part of some constant times $z^4,$ where $z = x + i y.$ I wrote it out,
$$z^4 = x^4 - 6 x^2 y^2 + y^4 +  4(x^3y - x y^3)i$$
Your initial polynomial is $1/4$ the imaginary part of this, so the conjugate can be
$$ \left(x^4 - 6 x^2 y^2 + y^4 \right) / 4  $$
