Calculating a harmonic conjugate Is the following reasoning correct?

Determine a harmonic conjugate to the function \begin{equation} f(x,y)=2y^{3}-6x^{2}y+4x^{2}-7xy-4y^{2}+3x+4y-4 \end{equation}

We first of all check if $f(x,y)$ is indeed a harmonic function. This amounts to show $f(x,y)$ satisfy the two-dimensional Laplace equation
\begin{equation}
\frac{\partial^{2 }f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}=0 \tag{1}
\end{equation}
We have $\frac{\partial^{2}f}{\partial x^{2}}=8-12y$ and $\frac{\partial^{2} f}{\partial y^{2}}=12y-8$. Thus, (1) is fulfilled, and so $f(x,y)$ is harmonic. 
Next, we seek to determine a harmonic conjugate to the given function. Let $u(x,y)=2y^{3}-6x^{2}y+4x^{2}-7xy-4y^{2}+3x+4y-4$. 
\begin{equation*}
u_{x}=v_{y} \iff -12xy+8x-7y+3=v_{y}
\end{equation*}
Integrate with respect to $y$
\begin{equation}
v=-6xy^{2}+8xy-\frac{7}{2}y^{2}+3y+h(x) \tag{2}
\end{equation}
where $h(x)$ is a function of $x$ alone. To determine this, we use the second Cauchy-Riemann equation $v_{x}=-u_{y}$
\begin{align*}
-u_{y}=v_{x} &\iff 6x^{2}+7x-6y^{2}+8y-4=h'(x)-6y^{2}+8y \\
&\iff h'(x)=6x^{2}+7x-4
\end{align*}
Integrating with respect to $x$ we have
\begin{equation}
h(x)=2x^{3}+\frac{7}{2}x^{2}-4x+C
\end{equation}
where $C$ is an arbitrary constant. Therefore, if we let $C=0$, then one harmonic conjugate of $u$ is given as:
\begin{equation}
v=2x^{3}+\frac{7}{2}x^{2}-6xy^{2}+8xy-4x-\frac{7}{2}y^{2}+3y
\end{equation}

Your reasoning is correct. I'll give an alternative solution, which works only for polynomials, but can be carried out mechanically: substitute $x=(z+\bar z)/2$ and $y=(z-\bar z)/(2i)$. Result: 
$$
f(z) = iz^3+ (2+7i/4)z^2+(3/2-2i)z -i\bar z^3 + (2-7i/4) \bar z^2 + (3/2+2i)\bar z - 4$$
The reason this polynomial is harmonic is that no monomial has $z$ and $\bar z$ together. The fact that we started with something real lends it certain symmetry: namely, the terms 
$$-i\bar z^3 + (2-7i/4) \bar z^2 + (3/2+2i)\bar z$$
are just the complex conjugates of $iz^3+ (2+7i/4)z^2+(3/2-2i)z$. Therefore, 
$$
f(z) = \operatorname{Re}(2iz^3+ 2(2+7i/4)z^2+2(3/2-2i)z -4)$$
and the imaginary part of same complex polynomial gives the harmonic conjugate of $f$. 
