So, here is the statement:
The temperature in a sheet in steady state satisfies the following differential equation
$$-\dfrac{\partial}{\partial x}\left(\alpha_x\dfrac{\partial u}{\partial x}\right)-\dfrac{\partial}{\partial y}\left(\alpha_y\dfrac{\partial u}{\partial y}\right) = f(x,y) \ \ \text{in} \ \ (x,y)\in\Omega$$
Consider the region $\Omega$ shown in the figure and suppose that the following boundary conditions are satisfied:
\begin{align*} u(x,y) & = 4 \ \ \text{for} \ (x,y) \in \Gamma_6 \ \text{and} \ (x,y) \in \Gamma_7 \\ \dfrac{\partial u}{\partial n}(x,y) & = x \ \ \text{for} \ (x,y) \in \Gamma_2 \ \text{and} \ (x,y) \in \Gamma_4 \\ \dfrac{\partial u}{\partial n}(x,y) & = y \ \ \text{for} \ (x,y) \in \Gamma_5 \\ \dfrac{\partial u}{\partial n}(x,y) & = \dfrac{x+y}{\sqrt{2}} \ \ \text{for} \ (x,y) \in \Gamma_1 \ \text{and} \ (x,y) \in \Gamma_3 \\ \end{align*}
where $\dfrac{\partial u}{\partial n}$ denotes the directional derivative in the direction to the normal at the boundary of the region $\Omega$ in the point $(x,y)$. Also sume that $\alpha_x=\alpha_y=10$ and $f(x,y)=\sin{(\pi x)}\cos{(\pi y)}$ in $\Omega$.
$R=5\ cm$
I've already set the PDE with the pdetool option in Matlab, giving it the form of $-\mathrm{div}(\mathrm{grad}\ u)$, but I'm completely lost when trying to set the Neumann's conditions with the directional derivatives.