In page $36$ of "Partial Differential Equation," Evan define $v(z) = \Phi(z-x) - \phi^x(z)$, where $\phi^x(z)$ is a corrector function, satisfying the following identities,
$$ \Delta \phi^x = 0 \ in \ U \tag{1}$$ $$ \phi^x = \Phi(y-x) \ in \ \partial U \tag{2}$$
$$\lim_{\epsilon \rightarrow 0} \int_{\partial B(x,\epsilon)} \frac{\partial v}{\partial \nu} w dS = \lim_{\epsilon \rightarrow 0} \int_{\partial B(x,\epsilon)} \frac{\partial \Phi}{\partial \nu}(x-z) w(z) dS $$
I am kind of lost about how he got the equation.
Note that you want to prove that $$\lim_{\varepsilon\to 0}\int_{\partial B(x,\varepsilon)}\frac{\partial \phi^x}{\partial \nu}wdS=0.$$ Note that $w$ is not a problematic function near $x$. In fact $w$ is harmonic function in $U\setminus\left\{y\right\}$. Thus it is smooth, hence bounded in the closed set $\partial B(x,\varepsilon)$ for $\varepsilon>0$ sufficiently small. The argument for $\phi^x$ is similar, due to being smooth in $U$, its derivatives are bounded on the closed set $\partial B(x,\varepsilon)$.