First, we find the characteristics along which $u$ is constant:

\begin{equation}

dt=\frac{dx}{2t}=\frac{dy}{x}

\end{equation}

Combining the first and second, we have $t^2=x+C_1$. So $x=t^2-C_1$. Subbing this in the last fraction and then combining with the first we obtain $t^3/3-C_1t=y+C_2$, or subbing in for $C_1$, $-2t^3/3+xt-y=C_2$. So our curve in $\mathbb{R}^3$ is described by the intersection of two surfaces, namely $t^2=x+C_1$ and $-2t^3/3+xt-y=C_2$. Since $u$ is constant along this curve, the general solution takes on the form of $u(x,y,t)=\phi(C_1,C_2)=\phi(t^2-x,-2t^3/3+xt-y)$. Now using the initial condition, we have $u|_{t=0}=\sin(x+y)=\phi(-x,-y)$. Thus we have $\phi(x,y)=\sin(-x-y)$. So we have the solution:

\begin{equation}

u(x,y,t)=\sin(x-t^2+2t^3/3-xt+y)

\end{equation}

A quick check shows that this is indeed a solution to the ODE and satisfies the initial conditions.