By Lagrange Multipliers, the function $f$ has no minima or maxima under constraint $g$? Find the extrema of $f$ subject to the stated condition.
$f(x,y)=x-y$ subject to $g(x,y)=x^2-y^2=2$.
Ok, by Lagrange Multiplier method, we find the points that satisfy $\nabla f(x,y) = \lambda \nabla g(x,y)$ for some $\lambda \in \mathbb R$.
Well then we have $(1,-1)=\lambda(2x,-2y)$ and the system: 
$$1=\lambda2x \\ -1 = -\lambda2y \implies 1 = \lambda2y$$
This also implies $\lambda,x,y \neq 0$. So we have $\lambda2x = \lambda2y \implies x = y$.
But $x=y \implies x^2-y^2=0$. And the constraint is that $x^2-y^2=2$. Does this mean that there cannot be any points that satisfy the system? And does it mean that there are no maxima/minima under these conditions? Please explain. 

As $x^2-y^2=(x-y)(x+y)$ the function becomes $(x,y)\mapsto2/(x+y)$, which clearly has no extrema.
