a. Euler-Lagrange equation as usual: $L=\frac{1}{2}(u_x^2+u_y^2-2yu)\implies L_{u_x}=u_x,\ L_{u_y}=u_y,\ L_u=-y$ and therefore
\begin{equation}
(u_x)_x+(yu_y)_y +u=0
\end{equation}
(I changed all signs).
b. Now
\begin{gather*}
\delta \Phi = \iint_\Omega \bigl[ u_x \delta u_x + yu_y\delta u_y-y\delta u\bigr]\,dxdy=\\
\iint\_\Omega \bigl[ -u_x x-( yu_y)_y -y\bigr] \delta u\,dxdy + \int_\Gamma \bigl[ u_x n_x + yu_y n_y\bigr]\delta u \,ds=\\
\int_\Gamma \bigl[ u_x n_x + yu_y n_y\bigr]\delta u\,ds
\end{gather*}
where $\Gamma$ is a boundary of $\Omega$ and $\mathbf{n}=(n_x,n_y)$ is an inner normal:
$\mathbf{n}=(1,0)$ at $\Gamma_1:=\{x=0\}$, $\mathbf{n}=(-1,0)$ at $\Gamma_3:=\{x=1\}$,
$\mathbf{n}=(0,1)$ at $\Gamma_2:=\{y=0\}$, $\mathbf{n}=(0,-1)$ at $\Gamma_4:=\{y=1\}$, and therefore
\begin{equation}
\delta \Phi = \int_{\Gamma_1} u_x\delta u\,dy - \int_{\Gamma_3} u_x\delta u\,dy -
\int_{\Gamma_4} u_y\delta u\,dx
\end{equation}
without any contribution from $\Gamma_2$. Since $\delta \Phi=0$ for all $\delta u$ (no constrain on the border) we get
\begin{equation}
u_x|_{x=0}=u_x|_{x=1}=u_y|_{y=1}=0
\end{equation}
and no condition at $y=0$.
