If I have a smooth real valued non-constant function $f$ with isolated minimum at $x_M$ does that mean that close to $x_M$ its second derivative must be positive? I.e. must there exist $a > 0$ so that
$$ \begin{align} f''(x_M + \varepsilon) > 0 \\ f''(x_M - \varepsilon) > 0 \end{align} $$
for all $\varepsilon \in (0, a]?$ If not can someone give an example?
No, nothing can be said about the sign of $f''$ in the vicinity of an isolated minimum.
An example is the function $f$ defined by $$ f(x) = e^{-1/x^2}\left(3 + 2 \sin(\frac {1}{x^2})\right) $$ for $x \ne 0$ and $f(0) = 0$. $f$ is infinitely often differentiable and has an absolute minimum at $x=0$. For $x \ne 0$ is $$ f''(x) = \frac{2e^{-1/x^2}}{x^6 } \left( 6 - 8 \cos(\frac {1}{x^2}) - 9x^2+6x^2\cos(\frac {1}{x^2})-6x^2\sin(\frac {1}{x^2})\right) \, . $$ For sufficiently small $|x|$ is $$ f''(x) \le \frac{2e^{-1/x^2}}{x^6 } \left( 7 - 8 \cos(\frac {1}{x^2})\right) $$ and that is negative if $\cos(1/x^2) = 1$, which happens arbitrarily close to zero.