Use the chain rule to determine $\frac{d}{dt}g(\mathbf{r}(t))$.

The functions $$\mathbf{r}:\mathbb{R}\mapsto \mathbb{R}^2$$ and $$g:\mathbb{R}^2\mapsto \mathbb{R}$$ are defined by $$\begin{equation*} \mathbf{r}(t) := \begin{bmatrix} \sinh{(t)} \\ t^2 \end{bmatrix} \quad \textit{and} \quad g(x,y) := x^2y. \end{equation*}$$
Use the chain rule to determine $$\frac{d}{dt}g(\mathbf{r}(t))$$.
Okay so $$g(\mathbf{r}(t)) = t^2\sinh^2{(t)}$$ so $$\frac{d}{dt}g(\mathbf{r}(t)) = 2t\sinh^2{(t)}+2t^2\cosh{(t)}$$. Is this correct? Thanks!

It is not correct and you also did not use Chain Rule. The second term is $$2t^{2} \sinh t\cosh t$$.
We have $$(g(r(t))'=g_x\frac {dx} {dt}+g_y \frac {dy} {dt}=2xy\frac {dx} {dt}+x^{2} \frac {dy} {dt}$$. Now verify that this gives the same answer.