Can we say that the point $(-3,2)$ is the local minimum?

For this graph can we say that the point $(-3,2)$ is the local minimum?

My attempt.

For any $x\in(-3-\delta,-3+\delta)$ there is no small $\epsilon>0$ , such that $\forall \delta\in(0,\epsilon)$ if $0<\delta<\epsilon$ then $f(x)>f(-3)$. So, this point is not a local minimum.

Is this attempt correct?

I'm afraid that you may misunderstand the definition of local minimum. I think the definition is $f(x_0)$ is a local minimum if there exists $\delta>0$ so that $\forall x\in(x_0-\delta,x_0+\delta)$, $f(x)\geq f(x_0)$. Therefore $(-3,2)$ is not a local minimum.

For a local minimum at a point $x=x_0$, the points in any neighborhood around this point, the function values must be larger than at the point $x_0$.

This is not the case at $x=-3$, where the function value is $2$. The points in any neighborhood around this point have lower function value.