Let $u(x,y,t)=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C(m,n,t)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}$ so that it automatically satisfies $u(0,y,t)=u(a,y,t)=u(x,0,t)=u(x,b,t)=0$ ,
Then $\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\dfrac{m^2\pi^2}{a^2}C(m,n,t)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}+\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\dfrac{n^2\pi^2}{b^2}C(m,n,t)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}+\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\dfrac{1}{c^2}\dfrac{\partial^2C(m,n,t)}{\partial t^2}\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}=0$
$\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\left(\dfrac{\partial^2C(m,n,t)}{\partial t^2}+\left(\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}\right)c^2\pi^2C(m,n,t)\right)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}=0$
$\therefore\dfrac{\partial^2C(m,n,t)}{\partial t^2}+\left(\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}\right)c^2\pi^2C(m,n,t)=0$
$C(m,n,t)=C_1(m,n)\sin\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)+C_2(m,n)\cos\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$\therefore u(x,y,t)=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_1(m,n)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}\sin\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$+\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_2(m,n)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}\cos\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$u(x,y,0)=0$ :
$\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_2(m,n)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}=0$
$C_2(m,n)=0$
$\therefore u(x,y,t)=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_1(m,n)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}\sin\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$u_t(x,y,t)=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_1(m,n)\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}\cos\left(\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi t\right)$
$u_t(x,y,0)=x(x-a)(y-b)$ :
$\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C_1(m,n)\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}c\pi\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}=x(x-a)(y-b)$
You need to handle extremely complicated double kernel inversion.