If [tex]h(x,y)=x^2-y^2[/tex], then you can find partial derivatives:
[tex] \frac{dh}{dx} =2x \\ \frac{dh}{dy} =-2y[/tex].
At point A(-3,2,5) (this is the point which lies on the surface, because [tex]5=(-3)^2-2^2=9-4=5[/tex]) these derivatives have values:
[tex] \frac{dh}{dx} |_A=2\cdot (-3)=-6 \\ \frac{dh}{dy} =-2\cdot 2=-4[/tex] -- the slopes at x and y- directions, respectively.
