The real part of $z=x+yi$ is given as $\Re f(z) = x^2-y^2+xy$. Therefore I have $$ \left\{ \begin{array}{c} G_x = 2x+y \\ G_y = -2y + x \end{array} \right. $$ $$ \left\{ \begin{array}{c} G_{xx} = 2 \\ G_{yy} = -2 \end{array} \right. $$ $G_{xx} + G_{yy} = 0$, therefore the function is harmonic.
Next step is $$ \left\{ \begin{array}{c} v_y = G_x = 2x+y \\ v_x = -G_y = 2y-x \end{array} \right. $$
To find the actual $v$, I integrate $v_y$ with respect to $y$: $$ v = \int_{y_0}^y (2x + \tau)d\tau + C(x) = 2\int_{y_0}^y xd\tau + \int_{y_0}^y \tau d\tau +C(x) = 2x\tau \Big|_{y_0}^y + {\frac{\tau^2}{2}}\Big|_{y_0}^y + C(x) = 2x(y-y_0) + {\frac{y^2}{2}} - {\frac{y_0^2}{2}} +C(x) $$
Let's calculate $v_x$: $$ v_x = 2x(y=y_0)_x + ({\frac{y^2}{2}} - {\frac{y_0^2}{2}})\prime + C\prime (x)$$ $$2(y-y_0) + y-y_0 + C\prime (x) = 2y -x$$ $$3y-3y_0+C\prime (x) = 2y-x$$ $$y-3y_0+C\prime (x)=-x$$ $$C\prime (x) = 3y_0-y-x$$ To find $C(x)$, we must integrate $C\prime (x)$: $$C(x) = \int_{x_0}^x (3y_0-y-\tau)d\tau = 3\int_{x_0}^x y_0d\tau - \int_{x_0}^x yd\tau - \int_{x_0}^x \tau d\tau = 3y_0\tau \Big|_{x_0}^x - y\tau \Big|_{x_0}^x - {\frac{\tau^2}{2}}\Big|_{x_0}^x = 3y_0(x-x_0) - y(x-x_0) - {\frac{x^2}{2}} + {\frac{x_0^2}{2}} = 3xy_0 - 3x_0y_0 - yx + yx_0 - {\frac{x^2}{2}} + {\frac{x_0^2}{2}}$$
Now we can have the full $v$: $$v = 2xy - 2xy_0 + {\frac{y^2}{2}} - {\frac{y_0^2}{2}} + 3xy_0 - 3x_0y_0 - yx + yx_0 - {\frac{x^2}{2}} + {\frac{x_0^2}{2}}$$
I'm really confused here. Were all my previous steps right? How do I get the final answer. By the way, it is $$ f(z) = (1-{\frac{i}{2}})z^2 + iC $$
Your answer is correct, but why have you done it so long. Consider $$v=\int v_y\ dy=2xy+\frac12y^2+h(x)+C$$ then $$v_x=2y+h'(x)=2y-x$$ and $h(x)=-\dfrac12x^2$. Finally $$f(z)=u+iv=(x^2-y^2+xy)-i\dfrac{1}{2}(x^2-y^2-4xy)+iC=(1-\frac12i)z^2+iC$$