General solution to Laplace Equation

You are asked to show that the general solution to an equation has a particular form, so you should start from the equation. You spotted correctly that it is a good idea to make a change of variables, $$z=x+iy, \qquad \bar{z} = x-iy.$$ By the Chain Rule, you can verify that $$\frac{\partial}{\partial x} = \frac{\partial}{\partial z} + \frac{\partial}{\partial \bar{z}} \quad \text{and} \quad \frac{\partial}{\partial y} = i\left( \frac{\partial}{\partial z} - \frac{\partial}{\partial \bar{z}} \right).$$ Let me also write $\phi(x,y) = \psi(z, \bar{z})$ for the new function that we want to solve for after making the change of variables. It is an easy calculation to check that $$\frac{\partial^2}{\partial x^2} \phi =\left(\frac{\partial}{\partial z} + \frac{\partial}{\partial \bar{z}} \right) \left( \frac{\partial}{\partial z} + \frac{\partial}{\partial \bar{z}} \right) \psi = \frac{\partial^2\psi}{\partial z^2} + 2 \frac{\partial^2 \psi}{\partial z \partial \bar{z}} + \frac{\partial^2 \psi}{\partial \bar{z}^2},$$ and that $$\frac{\partial^2}{\partial y^2} \phi =(i)^2 \left(\frac{\partial}{\partial z} - \frac{\partial}{\partial \bar{z}} \right) \left( \frac{\partial}{\partial z} - \frac{\partial}{\partial \bar{z}} \right) \psi = -\frac{\partial^2\psi}{\partial z^2} + 2 \frac{\partial^2 \psi}{\partial z \partial \bar{z}} - \frac{\partial^2 \psi}{\partial \bar{z}^2}.$$ So your equation becomes $$4 \frac{\partial^2 \psi}{\partial z \partial \bar{z}} = 0.$$ Can you see how to finish it off? [Hint: integrate].