''min $c^tx$ subject to $x^tAx=1$'': is is possible to solve it with Lagrange multiplier or in the scope of KKT? I find a problem:

Minimize $c^tx$ subject to $x^tAx=1$, where $A$ is a positive semidefinite symmetric matrix.

But the question obligates to use KKT but I am trying to apply simple Lagrange multiplier solution but cannot cope with it.
Is that question solvable via Lagrange Multiplier or this method is only available for linear equality constraints?
 A: If $A$ is a positive definite matrix then you can use usual Lagrangian methods but if $A$ is only positive semi-definite matrix then you have some constraint on the $x_s$ that is missing from your original question. For example, if $A=\left(\begin{array}{cc}0 & 0 \\0 & 1\end{array}\right)$ and $c=(-1,2)$ then there is no bounded solution. The "solution" in this example is $x_1=-\infty$ and $x_2=1$.
In what follows I assume $A$ is a positive definite matrix.
The first-order condition for the optimization problem give us:
$$c - \lambda\,\left(A^t+A\right)x=0 \Rightarrow x=\dfrac{\left(A^t+A\right)^{-1}c}{\lambda}\Rightarrow x^tAx=\dfrac{c^t \left(\left(A^t+A\right)^{-1}\right)^tA\left(A^t+A\right)^{-1}c}{\lambda^2}=\dfrac{c^t \left(A^t+A\right)^{-1}A\left(A^t+A\right)^{-1}c}{\lambda^2}=1$$ 
where $\left(A^t+A\right)^{-1}$ is the inverse matrix of $\left(A^t+A\right)$ (notice that the existence of $\left(A^t+A\right)^{-1}$ is not guaranteed if $A$ were to be only semi definite). So $\lambda=\sqrt{c^t \left(A^t+A\right)^{-1}A\left(A^t+A\right)^{-1}c}$ and we finally get: $$ \boxed{x=\dfrac{\left(A^t+A\right)^{-1}c}{\sqrt{c^t \left(A^t+A\right)^{-1}A\left(A^t+A\right)^{-1}c}}}.$$
Suggestion: don't get intimidated with vector calculus. It always helps to do simpler versions of the problem in $\mathbb{R}^2$ to check you are on the right track. 
Side work: $\nabla_x\left( c^t x\right) = c$ so $\nabla_x\left( y^tAx\right)= A^t y$ and $\nabla_y\left( y^tAx\right)= Ax$.
