Fortunately, there is a quick way to decide: we use Sylvester's Criterion, since the matrix is symmetric. It will be positive definite if and only if the principal minors are all positive.
One minor is $1 > 0$, ok. The second one is $\omega$. The last one is $-13\omega - 43^2-\pi^2\omega$.
If $\omega > 0$, the third minor is negative. So we can't have a local minimum.
If $\omega < 0$, then the first and the second minors will have opposite signs, so we can't have a local minimum either. If $\omega = 0$, the first and third minors will have opposite signs, problem again.
The matrix will be negative definite if its opposite is positive definite, so you repeat that analysis.
So we have neither maximum nor minimum.