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}
Best Answer
Yet another shortcut. Since $u$ is harmonic (on the simply connected domain $\mathbb{C}$), there has to be a harmonic conjugate $v$. Let $F = u+iv$ be the corresponding holomorphic function. It follows from (the derivation of) Cauchy-Riemann's equations that: $$ F' = u'_x - i\,u'_y = -12xy + 8x -7y + 3 + i(6x^2+7x-6y^2+8y-4). $$ Let $G(z) = 3 + 8z + i(6z^2+7z-4)$. Then $G(z) = F'(z)$ if $z$ is real, so by the identity theorem, $G = F'$ for all $z$. Hence $$ F(z) = 3z + 4z^2 - 4 + i(2z^3+\frac72z^2-4z+C) $$ for some real constant $C$ (the real part of the constant of integration has to be $4$ to match $u$). Finally $$ v = \operatorname{Im}(F(z)). $$