Surface $x^2+y^2-z^2=4$ - Tangent plane 
Give the equation of the tangent plan to the surface $x^2+y^2-z^2=4$
  where $(\hat{x},\hat{y})=(2,2)$ and $\hat{z}=-2$.

$(x^2+y^2)-4=z^2 \iff z=\sqrt{(x^2+y^2)-4}$ or $z=-\sqrt{(x^2+y^2)-4}$. As $\hat{z}$ is negative,then we have to take the first function.
$$\frac{\partial z}{\partial x}(\hat{x},\hat{y}) = \frac{-\hat{x}}{\sqrt{\hat{x}^2+\hat{y}^2-4}}$$ and $$\frac{\partial z}{\partial y}(\hat{x},\hat{y}) = \frac{-\hat{y}}{\sqrt{\hat{x}^2+\hat{y}^2-4}}$$
So $$\nabla f(\hat{x},\hat{y}) = \frac{-(\hat{x},\hat{y})}{\sqrt{\hat{x}^2+\hat{y}^2-4}}$$.
Hence we could define the plan as $z+2 = 2(x-2)+2(y-2)$
Am I right? Otherwise, could I modify something to get the right answer? Is there exist a shorter way to find this tangent plane? 

A shorter way is to define $f(x,y,z) = x^2+y^2 -z^2 - 4$ and note that $\nabla f(2,2,-2)$ is normal to the tangent plane at the point (if the gradient is not the zero vector, sure). One readily checks that $$\nabla f(2,2,-2) = (4,4,4),$$so one simpler normal vector is $(1,1,1)$. So the tangent plane is $$(x-2)+(y-2)+(z+2) = 0,$$that is: $$x+y+z = 2.$$
You got computations wrong. We have $$\frac{\partial z}{\partial x}(2,2) = \frac{\partial z}{\partial y}(2,2) = -1$$instead of $2$. And also in "we have to take the first function", you proabably meant "second", although you procceded correctly.
