Show that the vector field $X(x, y, z)=(xy-z^2, yz-x^2, x^2+z^2+xz-1)$ is tangent to the set $x^2 + y^2 + z^2 = 1$ I know I need to find functions $F(t)$, $G(t)$, and $H(t)$ such that $F(0)=x$, $G(0)=y$, and $H(0)=z$ and $F'(0)=xy-z^2$, $G'(0)=yz-x^2$, and $H'(0)=x^2+z^2+xz-1$.  It's also necessary that $(F(t))^2 +(G(t))^2 + (H(t))^2 = 1$ for all $t$. Just not sure where to go from here...
 A: It sounds like you've been taught more from a pure differential geometry perspective, taking about curves on manifolds and such.  I'll try to connect some of the disparate notions here into a coherent whole.
You've been given a set of position vectors ("points") that form a manifold.  Let's not worry about what a manifold is.  Let's just say it's a useful word that captures what is typically meant when talking about surfaces, volumes, and such, without necessarily choosing whether the object is 2d or 3d or whatever else.
At some position vector $p$, we can draw a curve $c(t)$ through the point, so that $c(0) = p$.  If the curve is smooth enough and differentiable, we can take a derivative $c'(0)$.  This is a vector, and a different kind of vector from a position vector.  This is usually called a tangent vector.
The set of all possible tangent vectors at a point $p$ forms what's called the tangent space at $p$.
Your definition of what it means for a vector field to be tangent is then equivalent to the following: 

A vector field $X$ is tangent to a manifold $A$ if at every point $p$, $X(p)$ lies entirely in the tangent space at $p$.

You should be able to see that this definition is equivalent to yours.  If $X(p)$ does indeed lie in the tangent space, than you can draw some curve whose derivative is the tangent vector $X(p)$.  We're not at all concerned with the behavior of the curve away from $p$, so finding these functions is overkill.
For your 2d manifold, the tangent space at any point $p$ is the plane tangent to the sphere at that point.  To say that $X(p)$ lies entirely in the tangent space is to say that $X(p)$ has no component perpendicular to the tangent plane, or rather, that it has no component along the normal to the tangent plane.  The normal vector could also be said to be $p$, which is exactly what MLT reasoned.  MLT showed that the projection of $X(p)$ along the normal direction $p$ is zero, therefore $X(p)$ has no normal component and must lie entirely in the tangent plane at each point $p$.
A: We will show that the vector field is perpendicular to any point $(x,y,z)$ on sphere by considering the dot product of the vector field and the normal to the sphere at that point.
$(xy-z^2,yz-x^2,x^2+z^2+xz-1)\cdot(x,y,z)=x^2y-xz^2+zy^2-yx^2+zx^2+z^3+xz^2-z=zy^2+zx^2+z^3-z=z(x^2+y^2+z^2-1).$ We know on the surface of the sphere $x^2+y^2+z^2-1=0$. Therefore the result holds.
