I cut and paste from Wikipedia:
For a function of more variables, one can look at the eigenvalues of the Hessian matrix at the critical point. The following test can be applied at a non-degenerate critical point (a, b, ...). If the Hessian is positive definite (equivalently, has all eigenvalues positive) at (a, b, ...), then f attains a local minimum at (a, b, ...). If the Hessian is negative definite (equivalently, has all eigenvalues negative) at (a, b, ...), then f attains a local maximum at (a, b, ...). If the Hessian has both positive and negative eigenvalues then (a, b, ...) is a saddle point for f (this is true even if (a, b, ...) is degenerate). Otherwise the test is inconclusive. Note that for functions of two or more variables, the determinant of the Hessian does not provide enough information to classify the critical point, because the number of jointly sufficient second-order conditions is equal to the number of variables, and the sign condition on the determinant of the Hessian is only one of the conditions. Note also that this statement of the second derivative test for many variables also applies in the two-variable and one-variable case. In the latter case, we recover the usual second derivative test.
See the original page here.
What you learned for the 2D case is a particular trick to test the positivity of the hessian matrix evaluated at a critical point. In higher dimension you must write down the hessian matrix and compute its signature, unless the specific structure of your function allows some easier approach.