suppose we have an objective function $f(x)$ that we want to maximize subject to constraint $g(x)=c$.
What I do next is set up Lagrangian optimization as follows and take the derivative
\begin{align*} L(x)&=f(x)+\lambda (g(x)-c)\\ \frac{dL(x)}{dx}&=f'(x)+\lambda g'(x) \end{align*}
So, we have to solve \begin{align*} f'(x)+\lambda g'(x)=0 \end{align*} in order to find a critical point. ( Up to here this was a usual procedure)
In my case the problem is that equation $f'(x)+\lambda g'(x)=0$ is way to complicated to solve. However, the good thing is that I have a candidate for the solution, call it $x^*$.
So since Lagrangian equation is sufficient and necessary for existence of critical point, I simply plugged in $x^*$
\begin{align*} f'(x^*)+\lambda g'(x^*)=0 \end{align*} and found required $\lambda$.
My question: is there a way to check whether $x^*$ is minimum or a maximum from what I did?
First, you need to make sure that $g'(x^*)$ fulfills a regularity condition (for instance: $g'(x^*)$ has full rank). Then you need to check second-order derivatives:
If the Hessian of the Lagrangian at $(x^*,\lambda)$ is positive (or negative) definite then $x^*$ is a local minimum (or maximum).
If the Hessian is not positive (or negative) semi-definite then $x^*$ is not a local minimum (or maximum).