$\nabla$$f(x,y,z)$ is the gradient of the function, which is a vector and gives the maximum rate of change of a function. If I find $\nabla$$f$ at a point, say at (x0,y0,z0), then is the vector that comes out relative to (x0,y0,z0), or is it relative to the origin.
That is, if $\nabla$$f$ at (x0,y0,z0) gave me a vector (a,b,c), would I need to subtract (x0,y0,z0) from (a,b,c) or is (a,b,c) the actual vector relative to the origin (namely, where is the tail of the vector (a,b,c)?)
If (a,b,c) came to be (1,2,3). If this Is relative to the origin, we have the vector as the line from (0,0,0) to (1,2,3). But this is a different vector compared to if the tail is at (x0,y0,z0) going to (1,2,3). Of course we can move the vector and it wont change, but which case is it?
The gradient gives the direction of fastest increase, so if you are at a point $(x_0,y_0,z_0)$, then to 'look' in the direction of (locally) steepest ascent you look at the points $(x_0,y_0,z_0) + t \nabla f(x_0,y_0,z_0)$ for $t \ge 0$.
So, in your example, you would look at the points $(x_0,y_0,z_0) + t (a,b,c)$ for $t \ge 0$.