Although your approach is not completely wrong, your approach is not the best.
Note that for $\nabla^2u(x,y)=0$ with conditions of the types $u(x,0)$ , $u(x,7)$ , $u(0,y)$ and $u(3,y)$ , according to http://eqworld.ipmnet.ru/en/solutions/lpde/lpde301.pdf#page=2 , we have special consideration:
$u(x,y)=\sum\limits_{n=1}^\infty A(n)\sinh\dfrac{n\pi(3-x)}{7}\sin\dfrac{n\pi y}{7}+\sum\limits_{n=1}^\infty B(n)\sinh\dfrac{n\pi x}{7}\sin\dfrac{n\pi y}{7}+\sum\limits_{n=1}^\infty C(n)\sin\dfrac{n\pi x}{3}\sinh\dfrac{n\pi(7-y)}{3}+\sum\limits_{n=1}^\infty D(n)\sin\dfrac{n\pi x}{3}\sinh\dfrac{n\pi y}{3}$
$u(0,y)=0$ :
$\sum\limits_{n=1}^\infty A(n)\sinh\dfrac{3n\pi}{7}\sin\dfrac{n\pi y}{7}=0$
$A(n)=0$
$\therefore u(x,y)=\sum\limits_{n=1}^\infty B(n)\sinh\dfrac{n\pi x}{7}\sin\dfrac{n\pi y}{7}+\sum\limits_{n=1}^\infty C(n)\sin\dfrac{n\pi x}{3}\sinh\dfrac{n\pi(7-y)}{3}+\sum\limits_{n=1}^\infty D(n)\sin\dfrac{n\pi x}{3}\sinh\dfrac{n\pi y}{3}$
$u(3,y)=0$ :
$\sum\limits_{n=1}^\infty B(n)\sinh\dfrac{3n\pi}{7}\sin\dfrac{n\pi y}{7}=0$
$B(n)=0$
$\therefore u(x,y)=\sum\limits_{n=1}^\infty C(n)\sin\dfrac{n\pi x}{3}\sinh\dfrac{n\pi(7-y)}{3}+\sum\limits_{n=1}^\infty D(n)\sin\dfrac{n\pi x}{3}\sinh\dfrac{n\pi y}{3}$
$u(x,0)=0$ :
$\sum\limits_{n=1}^\infty C(n)\sinh\dfrac{7n\pi}{3}\sin\dfrac{n\pi x}{3}=0$
$C(n)=0$
$\therefore u(x,y)=\sum\limits_{n=1}^\infty D(n)\sin\dfrac{n\pi x}{3}\sinh\dfrac{n\pi y}{3}$
$u(x,7)=\sin\dfrac{\pi x}{3}$ :
$\sum\limits_{n=1}^\infty D(n)\sinh\dfrac{7n\pi}{3}\sin\dfrac{n\pi x}{3}=\sin\dfrac{\pi x}{3}$
$D(n)=\begin{cases}\text{csch}\dfrac{7\pi}{3}&\text{when}~n=1\\0&\text{when}~n\neq1\end{cases}$
$\therefore u(x,y)=\text{csch}\dfrac{7\pi}{3}\sin\dfrac{\pi x}{3}\sinh\dfrac{\pi y}{3}$
Note that this solution suitable for $x,y\in\mathbb{C}$ , not only suitable for $0\leq x\leq3$ and $0 .