Minimum value

We note that $\color{#3D99F6}{y}$ increases with $\color{#3D99F6}{x^2+x+0.75}$ , therefore $\color{#3D99F6}{y}$ is minimum when $\color{#3D99F6}{x^2+x+0.75}$ is minimum.

$\begin{aligned} \color{#3D99F6}{y} & = \ln \left(\arcsin \left(\color{#3D99F6}{x^2+x+0.75}\right)\right) \\ & = \ln \left(\arcsin \left[\color{#3D99F6}{\left(x+\frac{1}{2}\right)^2}+\frac{1}{2}\right]\right) \quad \quad \small \color{#3D99F6}{\text{Since }\left(x+\frac{1}{2}\right)^2 \ge 0 \, \, \implies x^2+x+0.75 \ge \frac{1}{2}.} \\ \Rightarrow \color{#3D99F6}{y_{min}} & = \ln \left(\arcsin \left(\color{#3D99F6}{\frac{1}{2}}\right)\right) \\ & = \ln \left(\frac{\pi}{6}\right) \\ & \approx \boxed{-0.647} \end{aligned}$