Since the first part of the problem asked for the second derivative as a function of $t$, the solution by Isaac is the one that requires the least additional effort. However, note that $\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2=\sin^2 t+\cos^2 t =1.$
We recognize this curve as an ellipse. The full ellipse is traced out, starting at $(0,3)$, and moving clockwise, as $t$ ranges over the interval $[0,2\pi)$. Since $0$ is not in the domain of $t$, there is a hole in our ellipse, albeit a tiny one.
The bottom half of the ellipse is the part that is concave up. This is traced out as $t$ goes from $\pi/2$ to $3\pi/2$. The interval where we have concavity upwards is partly a matter of definition, either $\pi/2 \lt t \lt 3\pi/2$ or $\pi/2 \le t \le 3\pi/2$. So the familiar geometry of the ellipse provides a check on the parametric calculation.
Comment: As was pointed out, you had to calculate $\dfrac{d^2y}{dx^2}$ anyway, probably by computing $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ first, then $\dfrac{dy}{dx}$. Then you needed to do some further differentiation for the second derivative. That's good, one needs to have full control over these things.
However, from the ordinary equation of the ellipse, we can in one line note that $\frac{2x}{2^2} +\frac{2y}{3^2}\frac{dy}{dx}=0.$ It is now easy to find $\dfrac{dy}{dx}$ in terms of $x$ and $y$, and hence, by easy substitution, in terms of $t$.