In the book of Evans (PDE), page 22, they say that $$\Phi(\boldsymbol x)=-\frac{1}{2\pi}\log|\boldsymbol x|$$ is a solution of $\Delta u=0$ in $\mathbb R^2$. How is this possible ? $$\Phi(x,y)=-\frac{1}{4\pi}\ln(x^2+y^2),$$
and thus $$\Delta \Phi(x,y)=-\frac{1}{4\pi}\left(\frac{2x}{x^2+y^2}+\frac{2y}{x^2+y^2}\right)\neq 0.$$
Where is the problem here ?
The problem is that you (correctly) calculated $\partial_x\Phi+\partial_y\Phi$ instead of $\Delta\Phi=\partial^2_x\Phi+\partial^2_y\Phi$.