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. 

The Lagrangian method brings conditionally stationary points to the fore, if there are any. In this example there are none, as you have found out.
Now $x^2-y^2=2$ defines a hyperbola $\gamma$ with apexes at $(\pm\sqrt{2},0)$ and asymptotes $y=\pm x$. The function $f(x,y):=x-y$ essentially measures the distance from the point $(x,y)$ to the ascending asymptote of $\gamma$. This distance is monotonically increasing (resp. decreasing) on each of the two branches of $\gamma$, whence there can be no local maximum or minimum of $f$ on $\gamma$.
