Answer
Use $l(t)=0, u(t)=3^{1/t}$ in the Squeeze Theorem.
Work Step by Step
Because the $sine$ function is bounded by $-1$ and $1$ we have:
$$-1\leq\sin\left(\dfrac{1}{t}\right)\leq 1$$
As $\sin^2\left(\dfrac{1}{t}\right)\geq 0$, we have:
$$0\leq\sin\left(\dfrac{1}{t}\right)\leq 1$$
We multiply both sides by $3^{1/t}$ (the inequality's signs remain the same because $3^{1/t}>0$):
$$0\leq\sin\left(\dfrac{1}{t}\right)3^{1/t}\leq 3^{1/t}$$
Let's note:
$$l(t)=0, u(t)=3^{1/t}$$
When $t\rightarrow 0^-$, the limits of the functions $l$ and $u$ are:
$$\begin{align*}
\lim_{t\rightarrow 0^-}l(t)&=\lim_{t\rightarrow 0^-}0=0\\
\lim_{t\rightarrow 0^-}u(t)&=\lim_{t\rightarrow 0^-}(3^{1/t})=0.
\end{align*}$$
We have:
$$\begin{align*}
l(t)&\leq\sin^2\left(\dfrac{1}{t}\right)3^{1/t}\leq u(t)\\
\lim_{t\rightarrow 0^-}u(t)&=\lim_{t\rightarrow 0^-}u(t)=0.
\end{align*}$$
Using the Squeeze Theorem we obtain:
$$\lim_{t\rightarrow 0^-}\sin^2\left(\dfrac{1}{t}\right)3^{1/t}=0$$
