Answer
$$y'=\sin t+t\cos t$$
Work Step by Step
$$y=t\log_3(e^{(\sin t)(\ln3)})$$
We have $\log_3(e^{(\sin t)(\ln3)})=\log_3(e^{\ln3})^{\sin t}=\log_3(3)^{\sin t}=\sin t$
Therefore, $$y=t\sin t$$
The derivative of $y$ is: $$y'=\sin t+t(\sin t)'$$
$$y'=\sin t+t\cos t$$